Triangles - Congruence Theorems starts from a practical fact: a triangle with fixed sides cannot change shape, which is why bridges, roof trusses and towers are built from triangles. It defines congruent triangles through a correspondence of vertices, takes SAS as an axiom, and proves ASA, AAS, SSS and RHS from it. These rules prove the isosceles triangle theorems and many CPCT results. The chapter also teaches propositions and converses, why SSA fails in general and when it works, and why proofs must never rely on an inaccurate diagram.
Key Concepts
1. Why Triangles Are Rigid
Take three straight rods and join their ends with loose pins, so each joint can turn freely. The frame will not squash. Four rods joined the same way make a square that you can push into a slanted rhombus, with the same side lengths.
This is the property called rigidity. A triangle is rigid: once the lengths of its three sides are fixed, its three angles are fixed too, and so its whole shape is fixed. A quadrilateral (a 4-gon) is not rigid: the same four side lengths can make many different shapes with different angles.
| Frame | Sides fixed? | Can the shape change? | Why |
|---|---|---|---|
| Triangle (3 rods) | Yes | No | Three lengths allow only one shape of triangle |
| Quadrilateral (4 rods) | Yes | Yes | The angles can open and close while the sides stay the same |
| Quadrilateral with one diagonal rod | Yes | No | The diagonal splits it into two rigid triangles |
Uses of triangle rigidity. Engineers use triangles wherever a structure must hold its shape under load.
- Bridges, roof trusses and towers: the members form a chain of triangles, so the frame does not sag or sway in wind.
- Bicycle frames: two triangles keep the frame stiff without adding weight.
- Gates and shelves: a diagonal plank or a triangular bracket stops a rectangle from leaning into a parallelogram.
Why triangles give strength and stability. In a square frame a push can turn the joints and change the angles. In a triangle, changing any angle would need a side to stretch or shrink, and the rods resist that strongly. So three sides decide the triangle, a fact that becomes the SSS condition.
2. Congruence and Correspondence
Two figures are congruent if one can be placed exactly over the other so that they cover each other completely. Congruent figures have the same shape and the same size. Two coins of the same denomination are congruent. A photo and its enlargement have the same shape but different sizes, so they are not congruent.
For triangles, placing one over the other means matching each vertex of one with a vertex of the other. This matching is called a correspondence.
Definition. Two triangles are congruent if, under some correspondence of their vertices, all three pairs of corresponding sides are equal and all three pairs of corresponding angles are equal.
We write △ABC ≅ △PQR. The order of letters carries the correspondence:
- A ↔ P, B ↔ Q, C ↔ R (first letter matches first letter, and so on).
- Sides: AB = PQ, BC = QR, CA = RP.
- Angles: ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R.
So one congruence statement packs six equalities, and the letter order matters. If △ABC ≅ △PQR, the statement △ABC ≅ △QPR is generally false: it claims BC = PR, which need not be true.
How to find the correct order. Match the vertices where equal angles sit, or use the common letters of equal sides. Solved Example 2 shows the method.
CPCT. Once two triangles are proved congruent, every other pair of corresponding parts is equal. We cite this as CPCT: corresponding parts of congruent triangles (are equal). So proving congruence from three parts gives you the other three parts.
| Step | What you write |
|---|---|
| 1. Name the two triangles | In △ABD and △ACD |
| 2. List three equal parts, each with a reason | AB = AC (given), ∠BAD = ∠CAD (given), AD = AD (common) |
| 3. State the rule and the correspondence | ∴ △ABD ≅ △ACD (SAS) |
| 4. Use CPCT for what you need | ∴ BD = CD (CPCT) |
3. The SAS Congruence Axiom
Three well-chosen parts are enough to fix a triangle. The most basic rule is SAS.
SAS congruence axiom. If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the two triangles are congruent.
The included angle of two sides is the angle formed between them at their common vertex. In △ABC:
| Pair of sides | Included angle |
|---|---|
| AB and AC | ∠A |
| AB and BC | ∠B |
| BC and CA | ∠C |
Why it is an axiom. SAS is accepted as true without proof, in the same spirit as Euclid’s postulates. You can check it by construction: draw a side of 5 cm, make an angle of 40° at one end, and mark 7 cm along the other arm. Joining the free ends closes the triangle in only one way. SAS is the starting rule, and every other congruence rule in this chapter is proved from it.
Worked illustration. In △ABC and △DEF, AB = DE = 4 cm, ∠B = ∠E = 65° and BC = EF = 6 cm. The angle 65° lies between AB and BC in the first triangle and between DE and EF in the second. So △ABC ≅ △DEF by SAS, and by CPCT, AC = DF, ∠A = ∠D and ∠C = ∠F.
A trap. If AB = DE, BC = EF and ∠A = ∠D, SAS cannot be used, because ∠A lies between AB and AC, and AC is not given. This is the SSA pattern, which you meet in concept 8.
4. The SSS Congruence Condition
SSS congruence. If the three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.
SSS is the rigidity of a triangle written in the language of congruence. Three rods of fixed lengths can be joined into only one shape of triangle, so any two triangles with the same three side lengths must fit over each other.
Proof of SSS from SAS (outline). Let AB = PQ, BC = QR and CA = RP.
- On the side of BC opposite to A, draw a ray from B making ∠CBX = ∠Q, and mark D on it with BD = QP.
- In △DBC and △PQR: BD = QP, ∠DBC = ∠PQR, BC = QR. So △DBC ≅ △PQR (SAS). By CPCT, DC = PR, and ∠BDC = ∠P.
- Now DB = PQ = AB and DC = PR = AC. Join AD.
- In △ABD, AB = DB, so ∠BAD = ∠BDA (angles opposite equal sides). In △ACD, AC = DC, so ∠CAD = ∠CDA.
- Adding, ∠BAC = ∠BDC. This step assumes AD crosses BC between B and C. If AD passes through B or C, use only the other isosceles triangle; if AD meets BC extended, subtract the angles instead of adding.
- In △ABC and △DBC: AB = DB, ∠BAC = ∠BDC, AC = DC. So △ABC ≅ △DBC (SAS).
- Since △DBC ≅ △PQR, we get △ABC ≅ △PQR.
Step 4 uses the isosceles triangle theorem from concept 7. That theorem is proved from SAS alone, so there is no circular reasoning.
Typical SSS setting: a kite. In quadrilateral ABCD, AB = AD and CB = CD. The diagonal AC is common to △ABC and △ADC. So △ABC ≅ △ADC (SSS), and by CPCT ∠BAC = ∠DAC. The diagonal AC bisects ∠A (and also ∠C).
5. ASA and AAS Conditions
ASA congruence. If two angles and the included side of one triangle are equal to two angles and the included side of another triangle, the triangles are congruent. The included side of two angles is the side joining their vertices. In △ABC, the side included between ∠B and ∠C is BC.
Proof of ASA from SAS. Given ∠B = ∠E, BC = EF and ∠C = ∠F in △ABC and △DEF.
- Case AB = DE. Then AB = DE, ∠B = ∠E, BC = EF, so △ABC ≅ △DEF by SAS. Done.
- Case AB > DE. Mark P on BA with PB = DE. In △PBC and △DEF: PB = DE, ∠B = ∠E, BC = EF. So △PBC ≅ △DEF (SAS), and by CPCT ∠PCB = ∠DFE.
- But ∠DFE = ∠ACB (given). So ∠PCB = ∠ACB. P lies strictly between B and A, so ray CP lies inside ∠ACB and ∠PCB is smaller than ∠ACB. This is impossible. So AB > DE cannot happen.
- Case AB < DE is ruled out the same way, with a point marked on ED.
- Only AB = DE is possible, and case 1 gives congruence.
This is a proof by contradiction: assume the opposite, reach something impossible, and conclude the original claim.
AAS congruence. If two angles and a non-included side of one triangle are equal to the corresponding angles and side of another, the triangles are congruent.
Why AAS works. Suppose ∠A = ∠D, ∠B = ∠E and BC = EF. The angle sum of a triangle is 180°, so
∠C = 180° - ∠A - ∠B = 180° - ∠D - ∠E = ∠F.
Now ∠B = ∠E, BC = EF, ∠C = ∠F, which is ASA. So AAS is ASA in disguise.
Worked illustration. In △PQR and △XYZ, ∠P = ∠X = 50°, ∠Q = ∠Y = 70° and QR = YZ = 8 cm. QR is opposite ∠P and YZ is opposite ∠X, and QR is not between ∠P and ∠Q. So △PQR ≅ △XYZ (AAS). Also ∠R = ∠Z = 180° - 50° - 70° = 60°.
Why AAA fails. Three equal angles fix the shape only. Draw an equilateral triangle of side 3 cm and another of side 6 cm. Both have angles 60°, 60°, 60°, but one is twice as large. So AAA is not a congruence condition: at least one side must be equal.
6. The RHS Condition for Right Triangles
In a right-angled triangle, the side opposite the right angle is the hypotenuse. It is the longest side.
RHS congruence. If in two right-angled triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other, the triangles are congruent. (R = right angle, H = hypotenuse, S = side.)
Proof of RHS. Let △ABC be right-angled at B and △DEF be right-angled at E, with AC = DF and AB = DE.
- Extend CB beyond B to a point G with BG = EF.
- ∠ABG = 90°, since it forms a linear pair with ∠ABC = 90°. In △ABG and △DEF: AB = DE, ∠ABG = ∠DEF = 90°, BG = EF. So △ABG ≅ △DEF (SAS), and AG = DF (CPCT).
- So AG = DF = AC. In △AGC, AG = AC, hence ∠AGC = ∠ACG (angles opposite equal sides).
- In △ABG and △ABC: ∠AGB = ∠ACB, ∠ABG = ∠ABC = 90°, AB = AB (common). So △ABG ≅ △ABC (AAS).
- So △ABC ≅ △ABG ≅ △DEF.
A quicker check using Pythagoras. In the right triangles, BC2 = AC2 - AB2 and EF2 = DF2 - DE2. The right-hand sides are equal, so BC = EF, and SSS (or SAS with the right angle) finishes it.
Worked illustration. △ABC is right-angled at B and △PQR is right-angled at Q. AC = PR = 13 cm, AB = PQ = 5 cm. By RHS, △ABC ≅ △PQR. If BC = 12 cm (check: 52 + 122 = 25 + 144 = 169 = 132), then QR = 12 cm by CPCT.
7. Isosceles Triangle Theorems
A triangle with two equal sides is isosceles. The equal sides are the legs, the third side is the base, the angles at the ends of the base are the base angles, and the angle between the equal sides is the vertex angle.
Theorem 1. Angles opposite to equal sides of a triangle are equal.
Given: △ABC with AB = AC. To prove: ∠B = ∠C.
Proof 1 (with a bisector). Draw AD, the bisector of ∠A, meeting BC at D. In △BAD and △CAD: AB = AC (given), ∠BAD = ∠CAD (AD bisects ∠A), AD = AD (common). So △BAD ≅ △CAD (SAS), and ∠ABD = ∠ACD (CPCT), that is ∠B = ∠C.
Proof 2 (no construction). Compare △ABC with itself, read the other way round, △ACB. AB = AC, ∠A = ∠A, AC = AB. So △ABC ≅ △ACB (SAS). By CPCT the angle at B in the first matches the angle at C in the second: ∠B = ∠C.
Theorem 2 (converse). Sides opposite to equal angles of a triangle are equal.
Given: △ABC with ∠B = ∠C. To prove: AB = AC.
Proof. Compare △ABC and △ACB. ∠B = ∠C, BC = CB, ∠C = ∠B. So △ABC ≅ △ACB (ASA). By CPCT, AB = AC. (Another route: draw the bisector AD of ∠A; then △ABD ≅ △ACD by AAS.)
Results that follow.
- In an isosceles triangle with AB = AC, the bisector of the vertex angle ∠A bisects the base (BD = DC, from Proof 1 by CPCT) and is perpendicular to it (∠ADB = ∠ADC and they form a linear pair, so each is 90°).
- Each angle of an equilateral triangle is 60°. AB = AC gives ∠B = ∠C, and AB = BC gives ∠A = ∠C. So all three angles are equal, and 3 × ∠A = 180° gives ∠A = 60°.
- A triangle with all three angles equal is equilateral (Theorem 2, used twice).
Angle calculations. If AB = AC and the vertex angle is ∠A = x, then each base angle is (180° - x)/2. If a base angle is y, the vertex angle is 180° - 2y.
| Given | Base angles | Vertex angle |
|---|---|---|
| ∠A = 40°, AB = AC | (180° - 40°)/2 = 70° each | 40° |
| ∠B = 65°, AB = AC | 65° each | 180° - 130° = 50° |
| ∠A = 90°, AB = AC | 45° each | 90° |
| Equilateral | 60° | 60° |
8. Propositions, Converses, SSA and Diagram Accuracy
Proposition. A proposition is a statement that is either true or false. Many geometry results have the form “If P, then Q”. P is the hypothesis (what is given) and Q is the conclusion (what follows).
Converse. The converse of “If P, then Q” is “If Q, then P”: the hypothesis and conclusion swap places.
| Proposition | Converse | Is the converse true? |
|---|---|---|
| If two sides of a triangle are equal, the angles opposite them are equal. | If two angles of a triangle are equal, the sides opposite them are equal. | Yes (Theorem 2) |
| If two triangles are congruent, they have equal areas. | If two triangles have equal areas, they are congruent. | No |
| If a triangle is equilateral, each of its angles is 60°. | If each angle of a triangle is 60°, it is equilateral. | Yes |
Not every converse is true. A true proposition can have a false converse. To show a statement is false, one counterexample is enough. For the area converse: a right triangle with legs 6 cm and 4 cm and a right triangle with legs 8 cm and 3 cm both have area (1/2) × 6 × 4 = (1/2) × 8 × 3 = 12 cm2. Their hypotenuses are √52 cm and √73 cm, so they are different triangles. So a converse always needs its own proof.
Why SSA fails in general. SSA means two sides and a non-included angle are equal. Counterexample: draw AM perpendicular to a line, with AM = 12 cm, and mark B on the line with BM = 16 cm, so AB = √(144 + 256) = 20 cm. Mark C1 between B and M and C2 beyond M, each 5 cm from M, so AC1 = AC2 = √(144 + 25) = 13 cm. Triangles ABC1 and ABC2 share AB = 20 cm and ∠B, and both have AC = 13 cm, yet BC1 = 11 cm and BC2 = 21 cm. They are not congruent.
The compass picture. Fix ∠B and the side BA. The third vertex C lies on the other arm of ∠B, at distance AC from A, so C is where a circle with centre A and radius AC cuts that arm. If AC is shorter than AB, the point B is outside this circle, and the arm can cut the circle at two points, one point, or none. Two points means two different triangles: this is the ambiguous case.
When SSA does work. If the side opposite the given angle is at least as long as the other given side (AC ≥ AB above), then B is inside or on the circle, and the arm from B meets the circle at no more than one usable point. At most one triangle fits the data, so SSA gives congruence. Two important special cases:
- The given angle is a right angle. The side opposite it is the hypotenuse, the longest side. This is exactly RHS.
- The given angle is obtuse. The side opposite an obtuse angle is the longest side, so at most one triangle fits. (That fact is standard but outside this chapter’s listed syllabus, so treat this case as a check only.)
In exams. In a proof, the right-angle case is cited as RHS. Never write “SSA” as a reason. The general test only tells you how many triangles fit the data.
The role of diagram accuracy. A proof must rest on given facts and proved results, never on what a sketch seems to show. A famous false “proof” that every triangle is isosceles uses the point O where the bisector of ∠A meets the perpendicular bisector of BC, drops perpendiculars from O to AB and AC, and uses AAS and RHS to add up equal pieces into AB = AC. Each congruence step is valid. The error is in the picture: when AB ≠ AC, O lies outside the triangle and one foot of a perpendicular lands on an extended side, so one length is really a difference. Draw figures to scale and check where points fall before adding or subtracting lengths.
Formula and Theorem Sheet
| Result | Statement | How it is justified |
|---|---|---|
| CPCT | Corresponding parts of congruent triangles are equal | Definition of congruence |
| SAS | Two sides and the included angle equal ⇒ congruent | Axiom |
| ASA | Two angles and the included side equal ⇒ congruent | Proved from SAS by contradiction |
| AAS | Two angles and a corresponding non-included side equal ⇒ congruent | Angle sum property turns it into ASA |
| SSS | Three sides equal ⇒ congruent | Proved from SAS and the isosceles theorem |
| RHS | Right angle, hypotenuse and one side equal ⇒ congruent | Proved from SAS and AAS (or Pythagoras then SSS) |
| Isosceles theorem | AB = AC ⇒ ∠B = ∠C | SAS |
| Converse | ∠B = ∠C ⇒ AB = AC | ASA or AAS |
| Base angles | If AB = AC and ∠A = x, then ∠B = ∠C = (180° - x)/2 | Isosceles theorem and angle sum |
| Equilateral triangle | Each angle = 60°; three equal angles ⇒ equilateral | Isosceles theorem and its converse |
| Perpendicular bisector | A point equidistant from A and B lies on the perpendicular bisector of AB, and every point on it is equidistant | SSS one way, SAS the other |
| Angle bisector | A point on the bisector of an angle is equidistant from its arms, and conversely | AAS one way, RHS the other |
| AAA, SSA | Not congruence conditions in general | Counterexamples |
| SSA valid case | Side opposite the given angle ≥ the other given side (includes right and obtuse angles) | Only one triangle can be drawn; in proofs quote RHS, never SSA |
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Important Definitions
- Congruent triangles: two triangles for which some correspondence of vertices makes all three pairs of corresponding sides and all three pairs of corresponding angles equal.
- Correspondence: a one-to-one matching of the vertices of one triangle with the vertices of another, shown by the order of letters in △ABC ≅ △PQR.
- CPCT: corresponding parts of congruent triangles are equal.
- Included angle: the angle formed at the common vertex of two sides of a triangle.
- Axiom: a statement accepted as true without proof and used as a starting point. SAS congruence is taken as an axiom.
- Hypotenuse: the side opposite the right angle in a right-angled triangle; it is the longest side.
- Isosceles triangle: a triangle with at least two equal sides.
- Equilateral triangle: a triangle with all three sides equal.
- Base angles: the two angles of an isosceles triangle opposite the equal sides, at the ends of the base.
- Proposition: a statement that is either true or false.
- Converse: the statement formed by swapping the hypothesis and conclusion: “If Q, then P”.
- Counterexample: a single example that satisfies the hypothesis of a statement but not its conclusion, which proves the statement false.
Solved Examples
Example 1: Reading a congruence statement
Question. △ABC ≅ △PQR. In △ABC, AB = 6 cm, BC = 8 cm, ∠A = 70° and ∠B = 50°. Find PQ, QR, ∠P, ∠Q and ∠R.
Solution. The correspondence is A ↔ P, B ↔ Q, C ↔ R.
- PQ = AB = 6 cm and QR = BC = 8 cm.
- ∠P = ∠A = 70° and ∠Q = ∠B = 50°.
- ∠C = 180° - 70° - 50° = 60°, so ∠R = ∠C = 60°.
Example 2: Writing the correct correspondence
Question. In △ABC and △XYZ, AB = YZ, BC = ZX and CA = XY. Write the congruence with the correct order of vertices and name the rule.
Solution.
- AB and BC share B; YZ and ZX share Z. So B ↔ Z.
- AB matches YZ, and B ↔ Z, so A ↔ Y.
- The remaining vertices match: C ↔ X.
- All three sides are equal, so △ABC ≅ △YZX (SSS).
Check: AB ↔ YZ, BC ↔ ZX, CA ↔ XY. All three agree with the data.
Example 3: Segments that bisect each other
Question. Line segments AB and CD bisect each other at O. Prove that AC = BD and AC ∥ BD.
Solution. O is the midpoint of both segments, so OA = OB and OC = OD.
- In △AOC and △BOD: OA = OB (given), ∠AOC = ∠BOD (vertically opposite angles), OC = OD (given).
- ∴ △AOC ≅ △BOD (SAS).
- ∴ AC = BD (CPCT).
- Also ∠OAC = ∠OBD (CPCT). These are alternate angles made by the transversal AB with the lines AC and BD.
- Equal alternate angles mean the lines are parallel, so AC ∥ BD.
Example 4: The diagonal of a kite
Question. In quadrilateral ABCD, AB = AD and CB = CD. Prove that the diagonal AC bisects ∠A and ∠C. If ∠BAC = 35°, find ∠BAD.
Solution.
- In △ABC and △ADC: AB = AD (given), CB = CD (given), AC = AC (common).
- ∴ △ABC ≅ △ADC (SSS).
- ∴ ∠BAC = ∠DAC and ∠BCA = ∠DCA (CPCT). So AC bisects ∠A and ∠C.
- ∠BAD = ∠BAC + ∠DAC = 35° + 35° = 70°.
Example 5: A point equidistant from two points
Question. M is the midpoint of segment AB, and P is a point with PA = PB (P not on AB). Prove that PM ⊥ AB.
Solution.
- In △PMA and △PMB: PA = PB (given), MA = MB (M is the midpoint), PM = PM (common).
- ∴ △PMA ≅ △PMB (SSS), so ∠PMA = ∠PMB (CPCT).
- ∠PMA and ∠PMB form a linear pair, so ∠PMA + ∠PMB = 180°.
- ∴ 2∠PMA = 180°, ∠PMA = 90°, and PM ⊥ AB.
So every point equidistant from A and B lies on the perpendicular bisector of AB. The converse (every point on the perpendicular bisector is equidistant from A and B) is also true, by SAS.
Example 6: Opposite sides from two pairs of parallel lines
Question. Lines l and m are parallel, and lines p and q are parallel. Line p meets l at A and m at B; line q meets m at C and l at D, so that ABCD is a quadrilateral. Prove that △ABC ≅ △CDA.
Solution. Join the diagonal AC.
- AB lies on p and DC lies on q, with p ∥ q. With transversal AC, ∠BAC = ∠DCA (alternate angles).
- BC lies on m and AD lies on l, with l ∥ m. With transversal AC, ∠BCA = ∠DAC (alternate angles).
- AC = CA (common).
- In △ABC and △CDA: ∠BAC = ∠DCA, AC = CA, ∠BCA = ∠DAC.
- ∴ △ABC ≅ △CDA (ASA).
By CPCT, AB = CD and BC = DA. This is the first step towards the parallelogram theorems in the 4-gons (Quadrilaterals) chapter.
Example 7: A point on an angle bisector
Question. Ray AP bisects ∠BAC. From P, PM ⊥ AB and PN ⊥ AC, with M on AB and N on AC. Prove that PM = PN and AM = AN.
Solution.
- In △AMP and △ANP: ∠PAM = ∠PAN (AP bisects ∠A), ∠AMP = ∠ANP = 90°, AP = AP (common).
- AP is opposite the right angles, so it is a non-included side. ∴ △AMP ≅ △ANP (AAS).
- ∴ PM = PN and AM = AN (CPCT).
Example 8: Equal altitudes
Question. In △ABC, BE ⊥ AC and CF ⊥ AB, with E on AC and F on AB. If BE = CF, prove that AB = AC.
Solution.
- In △BCF and △CBE: ∠BFC = ∠CEB = 90°, hypotenuse BC = CB (common), CF = BE (given).
- ∴ △BCF ≅ △CBE (RHS).
- ∴ ∠FBC = ∠ECB (CPCT), that is ∠ABC = ∠ACB.
- Sides opposite equal angles are equal, so AC = AB.
Example 9: Angles in an isosceles triangle
Question. In △ABC, AB = AC. Side BC is extended beyond C to D, and ∠ACD = 115°. Find all three angles of △ABC.
Solution.
- ∠ACB and ∠ACD form a linear pair: ∠ACB = 180° - 115° = 65°.
- AB = AC, so ∠ABC = ∠ACB = 65°.
- ∠A = 180° - 65° - 65° = 50°.
Answer: ∠A = 50°, ∠B = 65°, ∠C = 65°.
Example 10: The right angle hidden in an extended isosceles triangle
Question. △ABC is isosceles with AB = AC. Side BA is extended beyond A to D such that AD = AB. Prove that ∠BCD = 90°.
Solution. AD = AB = AC.
- In △ABC, AB = AC, so ∠ABC = ∠ACB. Call each x.
- In △ACD, AC = AD, so ∠ACD = ∠ADC. Call each y.
- ∠BCD = ∠ACB + ∠ACD = x + y.
- Angle sum in △BCD: ∠DBC + ∠BCD + ∠CDB = 180°, so x + (x + y) + y = 180°.
- 2(x + y) = 180°, so x + y = 90°. ∴ ∠BCD = 90°.
Competency-Based Questions (with answers)
Q1. Case study: the wobbly gate
A carpenter builds a rectangular wooden gate by nailing four planks at the corners. After a few weeks the gate sags: its corners are no longer right angles, though every plank is the same length as before.
(a) Why did the shape change while the side lengths stayed the same?
(b) What single plank should the carpenter add, and why does it work?
(c) Which congruence condition expresses the property used in (b)?
Answer. (a) A 4-gon is not rigid. Four fixed side lengths allow many different sets of angles, so the frame can lean into a parallelogram. (b) Add one diagonal plank joining opposite corners. It divides the gate into two triangles, and each triangle with fixed sides has fixed angles, so the gate holds its shape. (c) SSS: three sides decide a triangle completely.
Q2. Case study: a roof truss
A roof truss is a triangle ABC with the ridge at A and the two ends of the tie beam at B and C. The two rafters AB and AC are cut to the same length. A post AD joins A to the midpoint D of BC.
(a) Prove that △ABD ≅ △ACD.
(b) Show that the post AD is perpendicular to the tie beam.
(c) If the rafters make an angle of 30° with the tie beam, find the angle between the two rafters at the ridge.
Answer. (a) AB = AC (equal rafters), BD = CD (D is the midpoint), AD = AD (common). So △ABD ≅ △ACD (SSS). (b) ∠ADB = ∠ADC by CPCT, and they form a linear pair, so each is 90°. (c) ∠B = ∠C = 30°, so ∠A = 180° - 60° = 120°.
Q3. Assertion and Reason
Assertion (A): If in △ABC and △PQR, AB = PQ, BC = QR and ∠A = ∠P, then the triangles must be congruent by SAS.
Reason (R): In the SAS rule, the equal angle must be the angle included between the two equal sides.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d). ∠A is not between AB and BC (the angle between them is ∠B), so the data is SSA and does not guarantee congruence. R correctly states the SAS requirement.
Q4. Assertion and Reason
Assertion (A): Two triangles with ∠A = ∠D, ∠B = ∠E and BC = EF are congruent.
Reason (R): If two angles of one triangle equal two angles of another, the third angles are also equal.
Answer: (a). Both are true. R gives ∠C = ∠F, which turns the given AAS data into ASA (∠B, BC, ∠C), and that explains A.
Q5. Assertion and Reason
Assertion (A): The statement “If all three angles of one triangle are equal to the three angles of another, the triangles are congruent” is false.
Reason (R): Two triangles can have the same angles but different sizes.
Answer: (a). Equilateral triangles of sides 3 cm and 6 cm have all angles 60° but are not congruent. R is the counterexample idea that explains A.
Q6. Source-based: proposition and converse
A student writes in her notebook: “Theorem: If a point lies on the bisector of an angle, it is equidistant from the arms of the angle. I think the converse is also true, but my teacher says I must prove it separately.”
(a) Write the converse.
(b) Why can the converse not be taken as true just because the theorem is true?
(c) Prove the converse, for a point P inside ∠BAC.
Answer. (a) If a point inside an angle is equidistant from its arms, it lies on the bisector of the angle. (b) A true proposition can have a false converse, so each converse needs its own proof. (c) Let PM ⊥ AB and PN ⊥ AC with PM = PN. In right triangles AMP and ANP, the hypotenuse AP is common and PM = PN. So △AMP ≅ △ANP (RHS), ∠PAM = ∠PAN (CPCT), and AP bisects ∠BAC.
Q7. Case study: two survey triangles
Two surveyors record the same data for two triangular plots ABC: AB = 20 m, the same ∠B, and AC = 13 m. The plots have different areas.
(a) Which pattern of data did the surveyors record?
(b) Explain how both plots can fit the same data.
(c) What extra measurement would settle the shape?
Answer. (a) SSA: two sides and an angle opposite one of them (∠B is opposite AC, not between AB and AC). (b) The side opposite ∠B (13 m) is shorter than AB (20 m). A circle of radius 13 m around A can cut the other arm of ∠B at two points, giving two different triangles. (c) Measuring the third side BC fixes the plot by SSS. Measuring ∠A would also work, since ∠A, AB and ∠B then fix it by ASA.
Q8. Reasoning: counting triangles
In △ABC, AB = 5 cm, ∠B = 100° and AC = 8 cm. How many non-congruent triangles fit this data? Give a reason.
Answer. Exactly one. The side opposite the given angle, AC = 8 cm, is longer than the other given side, AB = 5 cm. So B lies inside the circle of radius 8 cm around A, and the other arm of ∠B, starting from a point inside the circle, crosses the circle at exactly one point C. Only one triangle fits the data, so SSA gives a unique triangle in this case.
Important Questions for Exams
1 Mark Questions
Q1. If △DEF ≅ △LMN, which side of △LMN equals EF?
Answer: MN (E ↔ M, F ↔ N).
Q2. Which congruence rule is taken as an axiom?
Answer: SAS.
Q3. In △PQR, PQ = PR and ∠P = 40°. Find ∠Q.
Answer: ∠Q = (180° - 40°)/2 = 70°.
Q4. Is AAA a congruence condition? Give a reason in one line.
Answer: No. Equilateral triangles of sides 2 cm and 5 cm have equal angles but different sizes.
Q5. Write the converse of: “If a triangle is equilateral, then each of its angles is 60°.”
Answer: If each angle of a triangle is 60°, then the triangle is equilateral.
2 Marks Questions
Q6. Why is a triangle called rigid? Give one real-life use of this property.
Answer: Once the three side lengths of a triangle are fixed, its angles and shape are fixed, so it cannot be deformed; a quadrilateral with fixed sides can change shape. Bridges and roof trusses are built from triangles so they do not sag.
Q7. In △ABC and △PQR, AB = PQ and AC = PR. Which one more equality makes them congruent by SAS? Which would make them congruent by SSS?
Answer: SAS needs the included angle: ∠A = ∠P. SSS needs the third side: BC = QR.
Q8. In △ABC, ∠B = ∠C = 55°. Which sides are equal, and what is ∠A?
Answer: Sides opposite equal angles are equal: AC (opposite ∠B) = AB (opposite ∠C). ∠A = 180° - 110° = 70°.
3 Marks Questions
Q9. Prove that the angles opposite to equal sides of a triangle are equal.
Answer: Given △ABC with AB = AC. Draw the bisector AD of ∠A meeting BC at D. In △BAD and △CAD: AB = AC (given), ∠BAD = ∠CAD (construction), AD = AD (common). So △BAD ≅ △CAD (SAS). Hence ∠ABD = ∠ACD (CPCT), that is ∠B = ∠C.
Q10. In an isosceles triangle ABC with AB = AC, BD and CE are the medians to AC and AB (D is the midpoint of AC and E the midpoint of AB). Prove that BD = CE.
Answer: AB = AC, so AE = AB/2 = AC/2 = AD. In △ABD and △ACE: AB = AC (given), ∠A = ∠A (common), AD = AE (halves of equal sides). So △ABD ≅ △ACE (SAS), and BD = CE (CPCT).
Q11. ABC is an isosceles triangle with AB = AC. Altitudes BE and CF are drawn to AC and AB. Prove that BE = CF.
Answer: In △ABE and △ACF: ∠AEB = ∠AFC = 90°, ∠A = ∠A (common), AB = AC (given, the side opposite the right angle in each). So △ABE ≅ △ACF (AAS), and BE = CF (CPCT).
5 Marks Questions
Q12. State and prove the ASA congruence rule, assuming the SAS axiom.
Answer: Statement: two triangles are congruent if two angles and the included side of one are equal to two angles and the included side of the other. Proof: let ∠B = ∠E, BC = EF, ∠C = ∠F in △ABC and △DEF. If AB = DE, then SAS (AB, ∠B, BC) gives congruence. Suppose AB > DE. Take P on BA with PB = DE. Then △PBC ≅ △DEF (SAS: PB = DE, ∠B = ∠E, BC = EF), so ∠PCB = ∠DFE = ∠ACB. Since P lies between A and B, ∠PCB is part of ∠ACB and is smaller than it, a contradiction. The case AB < DE is ruled out by the same argument with a point on ED. So AB = DE and △ABC ≅ △DEF by SAS.
Q13. (a) Explain with a counterexample why SSA is not, in general, a congruence condition. (b) State one situation where SSA does guarantee congruence, and name the rule.
Answer: (a) Draw a line, a point M on it and AM ⊥ line with AM = 12 cm. Take B on the line with BM = 16 cm, so AB = √(144 + 256) = 20 cm. Take C1 between B and M and C2 beyond M, each 5 cm from M, so AC1 = AC2 = √(144 + 25) = 13 cm. △ABC1 and △ABC2 share AB = 20 cm and ∠B, and have AC = 13 cm, yet BC1 = 11 cm and BC2 = 21 cm. The triangles are not congruent, so SSA fails. (b) When the given angle is a right angle, the side opposite it is the hypotenuse and SSA becomes RHS, which guarantees congruence.
Q14. In △ABC, AB = AC. D is a point on BC such that BD = CD. (a) Prove that AD bisects ∠A. (b) Prove that AD ⊥ BC. (c) If ∠BAC = 80°, find ∠ABD and ∠BAD.
Answer: (a) In △ABD and △ACD: AB = AC (given), BD = CD (given), AD = AD (common). So △ABD ≅ △ACD (SSS), and ∠BAD = ∠CAD (CPCT). So AD bisects ∠A. (b) ∠ADB = ∠ADC (CPCT), and they form a linear pair, so each is 90°. So AD ⊥ BC. (c) ∠ABD = (180° - 80°)/2 = 50°, and ∠BAD = 80°/2 = 40°. Check in △ABD: 50° + 40° + 90° = 180°.
Common Mistakes and Examiner Tips
- Mistake: writing the congruence with vertices in the wrong order, such as △ABC ≅ △QPR when A matches P. Fix: find the matching vertices first from the equal angles or from the common letters of equal sides, then write the letters in that order.
- Mistake: using SAS with an angle that is not between the two sides. Fix: check that the angle’s vertex is the common letter of the two sides (for AB and BC, it must be ∠B).
- Mistake: quoting “SSA” or “ASS” as a congruence rule. Fix: only SAS, SSS, ASA, AAS and RHS are accepted. With two sides and a non-included angle, look for a right angle (RHS) or for more information.
- Mistake: using CPCT before congruence is proved. Fix: CPCT always comes after the line “∴ △___ ≅ △___ (rule)”.
- Mistake: in AAS, matching a side opposite one angle with a side opposite a different angle. Fix: the equal sides must be opposite the same pair of equal angles.
- Mistake: applying RHS to triangles without a right angle, or using a side that is not the hypotenuse. Fix: state the right angle and name the hypotenuse (the side opposite it) in your working.
- Mistake: listing parts without reasons. Fix: each equality needs its reason in brackets: given, common, vertically opposite angles, alternate angles, construction, midpoint, and so on.
- Mistake: assuming a converse is true because the theorem is true. Fix: test the converse. Prove it, or give one counterexample to show it is false.
- Mistake: reading equal lengths or right angles off a rough sketch. Fix: use only facts that are given or proved. Mark given equal parts on your figure with tick marks and arcs, and draw figures neatly enough that points appear where they really lie.
- Mistake: in isosceles angle problems, treating the vertex angle as a base angle. Fix: the base angles are opposite the equal sides. If AB = AC, the base angles are ∠B and ∠C, and ∠A is the vertex angle.
Quick Revision Points
- A triangle is rigid: fixed side lengths fix its angles and shape. A quadrilateral is not rigid.
- △ABC ≅ △PQR means A ↔ P, B ↔ Q, C ↔ R, with six pairs of equal parts.
- CPCT: corresponding parts of congruent triangles are equal. Use it only after proving congruence.
- SAS (two sides and the included angle) is taken as an axiom.
- ASA (two angles and the included side) is proved from SAS by contradiction.
- AAS works because equal pairs of angles force the third angles to be equal (angle sum 180°).
- RHS: right angle, hypotenuse and one side, for right-angled triangles only.
- AAA does not give congruence: equal angles, different sizes.
- SSA does not give congruence in general: the same data can fit two different triangles.
- SSA works when the side opposite the given angle is at least as long as the other given side, which covers right (RHS) and obtuse angles. In a proof, quote RHS, never SSA.
- Angles opposite equal sides are equal. Sides opposite equal angles are equal.
- With AB = AC and ∠A = x, each base angle is (180° - x)/2.
- Each angle of an equilateral triangle is 60°; a triangle with three equal angles is equilateral.
- Points equidistant from A and B lie on the perpendicular bisector of AB, and conversely.
- Points on an angle bisector are equidistant from the arms, and conversely.
- A true proposition can have a false converse; one counterexample disproves a statement.
- Never trust a sketch over the given facts: a wrongly placed point can “prove” that every triangle is isosceles.
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams: periodic tests, the half-yearly exam and the annual exam. It is also the foundation for the geometry proofs of Class 10 boards.
In the CBSE Class IX Mathematics course structure for 2026-27, this chapter sits in Unit IV, Geometry, with Euclid’s Geometry, Lines and Angles, 4-gons (Quadrilaterals) and Circles. Unit IV carries 25 marks out of the 80-mark theory paper and is allotted 69 teaching periods. The course structure gives marks for the unit as a whole and does not split them chapter by chapter. The remaining 20 marks are internal assessment (pen-paper tests and multiple assessment 10, portfolio 5, lab practical 5).
| Unit | Marks (out of 80) |
|---|---|
| I. Number System | 07 |
| II. Algebra | 20 |
| III. Coordinate Geometry | 04 |
| IV. Geometry | 25 |
| V. Mensuration | 14 |
| VI. Statistics and Probability | 10 |
The question paper design groups questions by thinking skill: remembering and understanding (43 marks, about 54%), applying (19 marks, about 24%), and analysing, evaluating and creating (18 marks, about 22%). From this chapter, stating rules and writing converses fit the first band, congruence proofs and isosceles angle work fit the second, and deciding whether data fixes a unique triangle or spotting a flawed proof fit the third.
Class 9 Maths – swipe through all 10 cards to understand the whole chapter.
Triangles Are Rigid
Fix the three sides of a triangle and its shape is fixed. A 4-gon with fixed sides can still change shape.
This is why bridges, trusses and towers are made of triangles.
- A square frame of hinged rods can be pushed into a rhombus.
- One diagonal brace splits a 4-gon into two rigid triangles.
- Rigidity written as a rule is the SSS condition.
Congruence and CPCT
Congruent triangles match in all six parts under a correspondence of vertices.
CPCT: corresponding parts of congruent triangles are equal.
- Sides: AB = PQ, BC = QR, CA = RP.
- Angles: ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R.
- Wrong letter order makes a wrong statement.
SAS Congruence
Two sides and the included angle equal means the triangles are congruent.
Taken as an axiom; all other rules are proved from it.
- The included angle sits at the common vertex of the two sides.
- For sides AB and BC, the included angle is ∠B.
- AB, BC and ∠A is SSA, and SAS cannot be used.
SSS Congruence
If all three sides of one triangle equal the three sides of another, the triangles are congruent.
Proved from SAS with the isosceles triangle theorem.
- Kite ABCD with AB = AD, CB = CD: △ABC ≅ △ADC.
- So the diagonal AC bisects ∠A and ∠C.
- PA = PB and M the midpoint of AB gives PM ⊥ AB.
ASA and AAS
Two angles and a corresponding side equal give congruence, whether or not the side is between them.
ASA is proved from SAS by contradiction.
- ASA: the side joins the vertices of the two angles.
- AAS: the side must be opposite the same angle in both.
- AAA fails: equal angles, different sizes.
RHS Congruence
Two right triangles with equal hypotenuses and one other pair of equal sides are congruent.
The hypotenuse is opposite the right angle and is the longest side.
- Only for right-angled triangles.
- Equal altitudes BE = CF in △ABC give AB = AC by RHS.
- Equal perpendiculars to the arms put a point on the angle bisector.
Isosceles Triangle Theorems
Angles opposite equal sides are equal, and sides opposite equal angles are equal.
Theorem by SAS, converse by ASA or AAS.
- AB = AC and ∠A = 40° gives ∠B = ∠C = 70°.
- Each angle of an equilateral triangle is 60°.
- The vertex-angle bisector bisects the base at right angles.
Propositions and Converses
The converse of ‘If P, then Q’ is ‘If Q, then P’. It must be proved on its own.
One counterexample is enough to show a statement is false.
- Isosceles theorem and its converse are both true.
- Congruent ⇒ equal areas is true; the converse is false.
- Equal perimeters do not force congruence: 3, 4, 5 and 4, 4, 4.
Why SSA Fails
Two sides and a non-included angle can fit two different triangles.
Valid when the side opposite the angle is at least as long as the other given side.
- The third vertex is where a circle cuts an arm: 0, 1 or 2 points.
- A right given angle turns SSA into RHS.
- In a proof, quote RHS; never write SSA as a reason.
Diagram Accuracy
A proof uses only given and proved facts, never what a rough sketch seems to show.
A misdrawn figure ‘proves’ every triangle is isosceles.
- In that false proof, the key point really lies outside the triangle.
- One length becomes a difference, so the addition step fails.
- Draw to scale and check where points actually fall.
📝 Practice Triangles – Congruence Theorems - 10 questions
Concept-check questions · with answers & explanations
Start →Close ✕
Why A: The statement matches A ↔ R, B ↔ P, C ↔ Q. So AB ↔ RP gives RP = 5 cm, and ∠C ↔ ∠Q gives ∠Q = 60°. Both parts of A are correct.
Why not B: AB matches RP, not PQ; PQ matches BC = 7 cm, and ∠P matches ∠B = 50°.
Why not C: ∠R = ∠A = 70° is right, but RQ matches AC, not BC. The 7 cm side BC matches PQ, and AC (opposite the 50° angle) is shorter than BC (opposite the 70° angle), so RQ is not 7 cm.
Why not D: PQ = 7 cm is right, but ∠P matches ∠B = 50°, not 60°.
Remember: Write the three pairs A ↔ R, B ↔ P, C ↔ Q before reading off any part.
Why B: AM = BM because M is the midpoint, the right angles at M are equal and lie between AM and PM, and PM is common. So SAS gives △PMA ≅ △PMB and PA = PB by CPCT, without assuming anything that is not given.
Why not A: It uses PA = PB, the very thing to be proved.
Why not C: RHS here needs the hypotenuses PA and PB to be equal, which again assumes the result.
Why not D: ∠PAM = ∠PBM is not given; it only follows after the congruence.
Remember: Use only given facts, shared sides and known angle facts; never the conclusion.
Why B: △AMP and △ANP are right-angled at M and N. Their hypotenuse AP is common and PM = PN is given. So RHS gives △AMP ≅ △ANP, and ∠PAM = ∠PAN by CPCT, which means AP bisects ∠BAC.
Why not A: The angle between PM and AP is ∠APM, which is not known to equal ∠APN, so SAS fails.
Why not C: Only one pair of angles, the right angles, is known; ASA needs two.
Why not D: AM = AN is not given; it only follows from the congruence.
Remember: Equal perpendicular distances from the arms put a point on the angle bisector, by RHS.
Why B: Step 1 (angles opposite equal sides in △ABC): let ∠ABC = ∠ACB = x. Step 2 (angles opposite equal sides in △ACD, since AD = AB = AC): let ∠ACD = ∠ADC = y. Step 3 (angle sum in △BCD): x + (x + y) + y = 180°, so 2(x + y) = 180° and ∠BCD = x + y = 90°.
Why not A: 60° would need the triangle to be equilateral, which is not given.
Why not C: 120° is 180° – 60°, from the same equilateral assumption.
Why not D: x and y change with ∠A, but their sum x + y is always 90°.
Remember: Name unknown angles x and y, then one angle-sum equation often cancels them.
Why C: Step 1 (30-60-90 triangle is half an equilateral triangle): the perpendicular AM from A to the other arm of ∠B satisfies AM = AB / 2 = 5 cm. Step 2 (circle of radius AC about A): since 5 cm is less than 6 cm and 6 cm is less than AB = 10 cm, the circle cuts the arm at two points, one on each side of M. So two different triangles are possible.
Why not A: No triangle would exist only if AC were less than 5 cm, the shortest distance from A to the arm.
Why not B: One triangle would need AC = 5 cm or AC of at least 10 cm; 6 cm gives two.
Why not D: A circle meets a straight arm in at most two points.
Remember: Angle opposite the shorter given side can mean two triangles.
Why B: A diagonal rod splits the frame into two triangles. Each triangle has three fixed side lengths, so it is rigid, and the whole frame can no longer change its angles. This is how gates and trusses are braced.
Why not A: A bigger square with four hinged rods is still a four-sided frame and still folds.
Why not C: The rods are already equal; a rhombus also has four equal sides, so equal lengths do not stop the change.
Why not D: A rhombus still has opposite sides parallel, so this condition allows the same wobble.
Remember: A diagonal brace turns one floppy 4-gon into two rigid triangles.
Why D: In △AOC and △BOD, OA = OB and OC = OD since O bisects both segments, and ∠AOC = ∠BOD as vertically opposite angles. The angle lies between the two sides, so the triangles are congruent by SAS and AC = BD by CPCT.
Why not A: △AOC and △AOD share OA, but ∠AOC and ∠AOD are a linear pair, not known to be equal.
Why not B: The conclusion is right but SSS would need AC = BD already, which is what we want to prove.
Why not C: No angle equality is available for ASA here, and AC = CD does not follow.
Remember: Crossing segments give vertically opposite angles, a ready-made included angle.
Why D: In △ABD and △ACD, ∠BAD = ∠CAD (AD bisects ∠A), AD is common, and ∠ADB = ∠ADC = 90° (AD ⊥ BC). The common side AD lies between these angles, so ASA gives △ABD ≅ △ACD and AB = AC by CPCT.
Why not A: BD and AD are sides of the same triangle and do not correspond to each other.
Why not B: BC is not a side of either smaller triangle, so it cannot come from this congruence.
Why not C: ∠B and ∠BAD are in the same triangle and add to 90°; they need not be equal.
Remember: Bisector plus perpendicular on the same segment: ASA, then the triangle is isosceles.
Why A: In △BAD and △CAD, AB = AC, ∠BAD = ∠CAD and AD is common, so they are congruent by SAS. By CPCT, BD = DC and ∠ADB = ∠ADC. These two angles form a linear pair, so each is 90° and AD ⊥ BC.
Why not B: ∠ADB and ∠ADC are equal and add to 180°, so each is 90°, not 60°.
Why not C: AD and BD are two sides of the same triangle; nothing makes them equal in general.
Why not D: In right △ABD, ∠B + ∠BAD = 90°; ∠B = 2∠BAD only for special shapes.
Remember: In an isosceles triangle the vertex bisector is the perpendicular bisector of the base.
Why C: If the given angle is 90°, the side opposite it is the hypotenuse, the longest side. A circle of that radius can meet the other arm at only one point on that side, so the triangle is fixed. This special SSA case is exactly the RHS rule.
Why not A: This is the case that can give two different triangles, so it fails.
Why not B: SSA does work in some cases, such as RHS or when the angle is opposite the longer given side.
Why not D: SSA can produce two triangles, so it is not always valid.
Remember: SSA is safe only when the angle faces the longer side, as in RHS.
Chapter Navigation
Previous: Sequences and Progressions Class 9 Notes
Related Chapters in Class 9 Maths
- Coordinate Geometry Class 9 Notes
- Introduction to Polynomials Class 9 Notes
- Number System Class 9 Notes
- Exploring Algebraic Identities Class 9 Notes
- Circles Class 9 Notes
- Area and Perimeter Class 9 Notes
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Frequently Asked Questions
Two triangles are congruent when some matching of their vertices makes all three pairs of corresponding sides and all three pairs of corresponding angles equal. Then one triangle fits exactly over the other. The statement △ABC ≅ △PQR records the matching A with P, B with Q and C with R, so the order of letters matters.
Some starting rule has to be accepted without proof, just as Euclid accepted his postulates. SAS is the natural choice because a construction with two sides and the angle between them closes in only one way. ASA is then proved from SAS by contradiction, AAS follows from ASA using the angle sum, and SSS and RHS are proved using SAS together with the isosceles triangle theorem.
CPCT stands for corresponding parts of congruent triangles. Once two triangles are proved congruent by SAS, ASA, AAS, SSS or RHS, every pair of corresponding sides and angles is equal. Use CPCT only after the congruence line has been written with its rule, and read the corresponding parts from the order of the letters in the statement.
With two sides and a non-included angle, the third vertex is where a circle cuts one arm of the angle, and that can happen at two points. For example, triangles with AB = 20 cm, AC = 13 cm and the same angle B can have BC = 11 cm or BC = 21 cm. Two different triangles fit the same data, so SSA does not guarantee congruence in general.
SSA works when the side opposite the given angle is at least as long as the other given side, because then only one triangle can be drawn. The case used in proofs is a right angle: the side opposite it is the hypotenuse, the longest side, and SSA becomes RHS. In a written proof, quote RHS and never SSA. The general test is for judging how many triangles fit the given data.
In ASA the equal side lies between the two equal angles, joining their vertices. In AAS the equal side is not between them. AAS is still valid because two equal pairs of angles force the third angles to be equal, since each triangle’s angles add to 180 degrees. That turns the AAS data into ASA. In AAS the equal sides must be opposite the same pair of equal angles.
Theorem 1: angles opposite to equal sides of a triangle are equal, proved by SAS using the bisector of the vertex angle. Theorem 2, its converse: sides opposite to equal angles are equal, proved by ASA or AAS. Together they show each angle of an equilateral triangle is 60 degrees and that the bisector of the vertex angle bisects the base at right angles.
No. The converse swaps the if part and the then part, and it needs its own proof. Congruent triangles have equal areas, but equal areas do not force congruence: right triangles with legs 6 cm and 4 cm, and with legs 8 cm and 3 cm, both have area 12 square cm, yet their hypotenuses are √52 cm and √73 cm. One counterexample like this is enough to show a converse is false.
