Triangles moves from congruence, where figures match in shape and size, to similarity, where only the shape must match. It defines similar polygons, proves the Basic Proportionality Theorem (Thales theorem) and its converse, and sets out the AAA, AA, SSS and SAS criteria for similar triangles. These results let you find a missing length from a parallel line, prove that angles or ratios are equal, and measure heights and distances indirectly from shadows. The chapter trains careful proof writing, with a reason for every step.
Key Concepts
1. Similar Figures: Same Shape, Any Size
In Class 9 you learnt that two figures are congruent if they have the same shape and the same size. This chapter widens the idea. Two figures having the same shape but not necessarily the same size are called similar figures.
Some families of figures are always similar, because every member has the same shape:
- All circles are similar. Circles with the same radius are congruent; circles with different radii are still similar.
- All squares are similar.
- All equilateral triangles are similar.
A circle and a square can never be similar, and neither can a triangle and a square, because their shapes differ.
From this, NCERT draws a key fact: all congruent figures are similar, but similar figures need not be congruent. Congruence is the special case of similarity where the size also matches.
Everyday picture. Stamp-size, passport-size and postcard-size prints of one photograph of the Taj Mahal are similar: every length is enlarged in the same ratio and every angle stays the same. Two same-size photographs of one person at age 10 and at age 40 are not similar, because the shape has changed.
2. Similar Polygons and the Scale Factor
To decide whether two figures are similar we need a precise rule. NCERT gives it for polygons:
Two polygons of the same number of sides are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).
For quadrilaterals ABCD and PQRS, similarity means ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R, ∠D = ∠S and AB/PQ = BC/QR = CD/RS = DA/SP.
The common ratio of corresponding sides is called the scale factor (or the Representative Fraction). World maps and blueprints of buildings are drawn using a suitable scale factor.
Activity 1. A cardboard quadrilateral ABCD held parallel to the table under a bulb at O casts a shadow A′B′C′D′. Light travels in straight lines, so A′ lies on ray OA, B′ on ray OB, and so on. The shadow is an enlargement: corresponding angles are equal and corresponding sides are in the same ratio, so the two quadrilaterals are similar.
Why both conditions are needed for polygons
| Pair of quadrilaterals | Angles equal? | Sides in the same ratio? | Similar? |
|---|---|---|---|
| A square and a rectangle (non-square) | Yes, all 90° | No | No |
| A square and a rhombus (non-square) | No | Yes, all sides of each are equal | No |
| Two squares of sides 3 cm and 5 cm | Yes | Yes, each ratio 3/5 | Yes |
So for polygons with four or more sides, either condition alone is not sufficient. You must check both.
Remark (NCERT): if one polygon is similar to a second polygon, and the second is similar to a third, then the first polygon is similar to the third.
3. Similarity of Triangles and the Correspondence of Vertices
A triangle is a polygon, so the same definition applies. Two triangles are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).
In ΔABC and ΔDEF, if ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB/DE = BC/EF = CA/FD, the triangles are similar. We write ΔABC ~ ΔDEF and read it as “triangle ABC is similar to triangle DEF”. The symbol ~ means “is similar to”, just as ≅ means “is congruent to”.
The order of letters carries meaning. Here A corresponds to D, B to E and C to F. For these triangles we cannot write ΔABC ~ ΔEDF or ΔABC ~ ΔFED, but we can write ΔBAC ~ ΔEDF, because the pairs (B, E), (A, D) and (C, F) still match.
Once the statement is written correctly, you can read every equal angle and every ratio straight off it: first letter with first letter, first two letters with first two letters, and so on.
Triangles whose corresponding angles are equal are called equiangular triangles. The Greek mathematician Thales (640 to 546 B.C.) stated that the ratio of any two corresponding sides in two equiangular triangles is always the same. He is believed to have used the result now called the Basic Proportionality Theorem, or Thales Theorem.
4. The Basic Proportionality Theorem (Theorem 6.1)
Activity 2. Draw an angle XAY. On arm AX mark five points P, Q, D, R and B so that AP = PQ = QD = DR = RB. Through B draw any line meeting arm AY at C. Through D draw a line parallel to BC meeting AC at E. D is the third of the five equal steps, so AD/DB = 3/2. On measuring, AE/EC also comes out as 3/2. This is the Basic Proportionality Theorem at work.
Theorem 6.1 (Basic Proportionality Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Proof (write this in full for a long answer)
Given: In ΔABC, a line parallel to side BC meets AB at D and AC at E, so DE || BC.
To prove: AD/DB = AE/EC.
Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.
Proof:
- Area of a triangle = ½ × base × height. Taking AD as base and EN as height, ar(ADE) = ½ AD × EN.
- Similarly ar(BDE) = ½ DB × EN, ar(ADE) = ½ AE × DM and ar(DEC) = ½ EC × DM.
- So ar(ADE)/ar(BDE) = (½ AD × EN)/(½ DB × EN) = AD/DB … (1)
- And ar(ADE)/ar(DEC) = (½ AE × DM)/(½ EC × DM) = AE/EC … (2)
- ΔBDE and ΔDEC are on the same base DE and between the same parallels BC and DE, so ar(BDE) = ar(DEC) … (3)
- From (1), (2) and (3): AD/DB = AE/EC. Hence proved.
Other forms of BPT (NCERT Example 1)
If DE || BC in ΔABC with D on AB and E on AC, then all of these hold:
| Form | Ratio | Type |
|---|---|---|
| Basic form | AD/DB = AE/EC | part / part |
| Inverted | DB/AD = EC/AE | part / part |
| Top part to whole side | AD/AB = AE/AC | part / whole |
| Bottom part to whole side | DB/AB = EC/AC | part / whole |
The rule: use the same kind of ratio on both sides. Part over part equals part over part; part over whole equals part over whole. Never set AD/DB equal to AE/AC.
Worked illustration
In ΔABC, DE || BC, AD = 4 cm, DB = 6 cm and AC = 15 cm. Find AE.
Use the part-to-whole form: AD/AB = AE/AC. AB = 4 + 6 = 10 cm, so 4/10 = AE/15, giving AE = 15 × 4/10 = 6 cm. Check: EC = 9 cm, and AD/DB = 4/6 = 2/3 = 6/9 = AE/EC.
5. The Converse of BPT (Theorem 6.2)
Activity 3. Draw an angle XAY. On ray AX mark B1, B2, B3, B4 and B so that AB1 = B1B2 = B2B3 = B3B4 = B4B. On ray AY mark C1, C2, C3, C4 and C in the same way with equal steps. Join B1C1 and BC. Then AB1/B1B = AC1/C1C = 1/4, and B1C1 turns out parallel to BC. Joining B2C2, B3C3 and B4C4 gives ratios 2/3, 3/2 and 4/1 on both arms, and each of these lines is also parallel to BC.
Theorem 6.2 (Converse of BPT): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Proof outline (by contradiction)
- Take D on AB and E on AC with AD/DB = AE/EC, and suppose DE is not parallel to BC.
- Draw DE′ parallel to BC, meeting AC at E′.
- By BPT, AD/DB = AE′/E′C. So AE/EC = AE′/E′C.
- Add 1 to both sides: (AE + EC)/EC = (AE′ + E′C)/E′C, that is AC/EC = AC/E′C.
- So EC = E′C, which means E and E′ are the same point. Hence DE is the line DE′, and DE || BC.
How to use the converse
BPT goes from parallel ⇒ equal ratios. The converse goes from equal ratios ⇒ parallel. To test whether a line is parallel to the third side, work out both ratios exactly (as fractions). If they are equal, the line is parallel. If they are unequal, the line is not parallel, because a parallel line would force equal ratios by BPT.
The converse also works with the part-to-whole form: if AD/AB = AE/AC, then DE || BC.
Mid-point theorem as a special case. If D and E are the mid-points of AB and AC, then AD/DB = 1 = AE/EC, so DE || BC by the converse (NCERT Exercise 6.2, Q8). In the other direction, a line through the mid-point D of AB parallel to BC gives AE/EC = AD/DB = 1, so it bisects AC (Exercise 6.2, Q7).
6. AAA and AA Similarity Criteria (Theorem 6.3)
Checking all three angles and all three ratios every time would be slow. As with congruence, a few pairs of parts are enough.
Activity 4. Draw BC = 3 cm and EF = 5 cm, and at both ends make angles of 60° and 40°, so the two triangles ABC and DEF have equal corresponding angles. BC/EF = 0.6, and on measuring, AB/DE and CA/FD also come out as about 0.6.
Theorem 6.3 (AAA criterion): If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.
Proof idea. Cut DP = AB and DQ = AC on DE and DF and join PQ. ΔABC ≅ ΔDPQ (SAS congruence), so ∠P = ∠B = ∠E and PQ || EF. BPT in ΔDEF then gives AB/DE = AC/DF, and in the same way AB/DE = BC/EF.
The AA criterion
By the angle sum property, if two angles of one triangle are equal to two angles of another, the third angles are also equal (each is 180° minus the same sum). So NCERT restates AAA as: if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. This is the AA similarity criterion, and it is the one you will use most in proofs.
Where equal angles come from in a figure:
- A common angle shared by both triangles (a vertex that belongs to both).
- Vertically opposite angles where two lines cross.
- Alternate or corresponding angles formed by parallel lines.
- Right angles, for example at the foot of an altitude or where a pole stands on the ground.
- Angles given equal in the question.
Illustration. ΔABC has ∠A = 50° and ∠B = 70°. ΔPQR has ∠Q = 70° and ∠R = 60°. Is ΔABC ~ ΔPQR? ∠C = 180° − 50° − 70° = 60° and ∠P = 180° − 70° − 60° = 50°. So ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R and ΔABC ~ ΔPQR by AA.
7. SSS and SAS Similarity Criteria (Theorems 6.4 and 6.5)
SSS criterion
Activity 5. Draw ΔABC with AB = 3 cm, BC = 6 cm, CA = 8 cm and ΔDEF with DE = 4.5 cm, EF = 9 cm, FD = 12 cm. Every ratio AB/DE, BC/EF, CA/FD equals 2/3. On measuring, ∠A = ∠D, ∠B = ∠E and ∠C = ∠F.
Theorem 6.4 (SSS criterion): If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.
Proof idea. Cut DP = AB and DQ = AC and join PQ. The converse of BPT gives PQ || EF, so ΔDPQ ~ ΔDEF and PQ/EF = DP/DE = BC/EF, which means PQ = BC. Then ΔABC ≅ ΔDPQ (SSS congruence), and the angles match.
Matching sides correctly. Sort each triangle’s sides from shortest to longest and pair them in that order. Example: ΔABC has AB = 2 cm, BC = 2.5 cm, CA = 3 cm, and ΔPQR has PQ = 6 cm, QR = 4 cm, RP = 5 cm. Sorted pairs: AB with QR (2/4), BC with RP (2.5/5), CA with PQ (3/6), every ratio 1/2. AB ↔ QR and BC ↔ RP share B and R, so B ↔ R, A ↔ Q and C ↔ P. The statement is ΔABC ~ ΔQRP, and so ∠A = ∠Q, ∠B = ∠R, ∠C = ∠P.
Key remark. For triangles, Theorems 6.3 and 6.4 show that one condition implies the other: equal angles force proportional sides, and proportional sides force equal angles. For polygons with four or more sides you must check both.
SAS criterion
Activity 6. Draw ΔABC with AB = 2 cm, ∠A = 50°, AC = 4 cm and ΔDEF with DE = 3 cm, ∠D = 50°, DF = 6 cm. Then AB/DE = AC/DF = 2/3, and ∠A (between AB and AC) equals ∠D (between DE and DF). On measuring, ∠B = ∠E and ∠C = ∠F, so ΔABC ~ ΔDEF.
Theorem 6.5 (SAS criterion): If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
Proof idea. Cut DP = AB and DQ = AC and join PQ. Then PQ || EF (converse of BPT) and ΔABC ≅ ΔDPQ (SAS congruence), so all three pairs of angles are equal and AAA applies.
The angle must be the included angle, the one between the two proportional sides. An equal angle somewhere else does not satisfy SAS.
RHS similarity (NCERT note to the reader)
If in two right triangles, the hypotenuse and one side of one triangle are proportional to the hypotenuse and one side of the other triangle, then the two triangles are similar. This may be called the RHS similarity criterion.
Choosing the right criterion
| What the question gives you | Criterion to use | Similarity condition to state |
|---|---|---|
| Two pairs of equal angles | AA | ∠A = ∠D and ∠B = ∠E |
| Three side lengths of each triangle | SSS | AB/DE = BC/EF = CA/FD |
| One equal angle and the two sides around it | SAS | ∠A = ∠D and AB/DE = AC/DF |
| Two right triangles with hypotenuse and a side | RHS | right angles, and hypotenuse/hypotenuse = side/side |
8. Using Similarity: Heights, Shadows and Other Lengths
Heights of mountains such as Mount Everest and distances of far objects such as the moon were not measured with a tape. They were found by indirect measurement, which is based on similarity: a small triangle that you can measure is similar to a large triangle that you cannot, so the unknown length follows from equal ratios.
Shadows at the same moment
At a given moment the sun’s rays fall on nearby objects at the same angle. A vertical pole and a vertical tower both make 90° with the ground. So the triangle formed by the pole, its shadow and the sun’s ray is similar (AA) to the triangle formed by the tower, its shadow and the ray:
height of pole / length of its shadow = height of tower / length of its shadow
The lamp-post problem
A person of height h stands at distance d from a lamp-post of height H, and the shadow has length x. The lamp, the top of the person’s head and the tip of the shadow lie on one straight line. Both triangles share the angle at the tip of the shadow and have a right angle at the ground, so they are similar by AA:
(d + x)/x = H/h
Notice that the big triangle’s base is the whole distance from the foot of the lamp-post to the tip of the shadow, which is d + x, and the small triangle’s base is only x.
Other lengths. In similar triangles, corresponding medians and angle bisectors are in the ratio of corresponding sides (Exercise 6.3, Q10 and Q16). If D on BC satisfies ∠ADC = ∠BAC, then CA2 = CB · CD (Solved Example 11).
Formula and Theorem Sheet
| Result | Statement | Symbols |
|---|---|---|
| Theorem 6.1 (BPT) | A line parallel to one side of a triangle divides the other two sides in the same ratio | DE || BC ⇒ AD/DB = AE/EC |
| Theorem 6.2 (Converse of BPT) | A line dividing two sides of a triangle in the same ratio is parallel to the third side | AD/DB = AE/EC ⇒ DE || BC |
| Theorem 6.3 (AAA) and AA criterion | Equal corresponding angles ⇒ similar; two pairs of equal angles are enough | ∠A = ∠D, ∠B = ∠E |
| Theorem 6.4 (SSS) | Sides proportional ⇒ angles equal ⇒ similar | AB/DE = BC/EF = CA/FD |
| Theorem 6.5 (SAS) | One angle equal and the sides including it proportional ⇒ similar | ∠A = ∠D, AB/DE = AC/DF |
| RHS similarity | Right triangles with hypotenuse and one side proportional are similar | hyp/hyp = side/side |
| Diagonals of a trapezium | Exercise 6.2 Q9 and Exercise 6.3 Q3 | AO/BO = CO/DO |
| Point D on BC with ∠ADC = ∠BAC | Exercise 6.3 Q13 | CA2 = CB · CD |
| Shadows | AA similarity with the sun’s rays | h1/s1 = h2/s2 |
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Important Definitions
- Congruent figures: figures that have the same shape and the same size.
- Similar figures: figures that have the same shape but not necessarily the same size.
- Similar polygons: two polygons of the same number of sides whose corresponding angles are equal and whose corresponding sides are in the same ratio (proportion).
- Scale factor (Representative Fraction): the common ratio of the corresponding sides of two similar polygons.
- Similar triangles: two triangles whose corresponding angles are equal and whose corresponding sides are in the same ratio. Written ΔABC ~ ΔDEF with the vertices in corresponding order.
- Equiangular triangles: two triangles whose corresponding angles are equal.
- RHS similarity criterion: if in two right triangles the hypotenuse and one side of one triangle are proportional to the hypotenuse and one side of the other, the triangles are similar.
- Indirect measurement: finding a height or distance that cannot be measured directly, by using similar figures.
Solved Examples (NCERT-Based)
Example 1: The part-to-whole form of BPT (NCERT Example 1)
If a line intersects sides AB and AC of a ΔABC at D and E respectively and is parallel to BC, prove that AD/AB = AE/AC.
Solution: DE || BC (given). So AD/DB = AE/EC (Theorem 6.1).
Inverting both sides, DB/AD = EC/AE. Adding 1 to both sides: DB/AD + 1 = EC/AE + 1, so (DB + AD)/AD = (EC + AE)/AE.
Since DB + AD = AB and EC + AE = AC, this gives AB/AD = AC/AE. Inverting again, AD/AB = AE/AC. Hence proved.
Example 2: Is EF parallel to QR? (NCERT Exercise 6.2, Q2)
E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR: (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Solution:
(i) PE/EQ = 3.9/3 = 1.3 and PF/FR = 3.6/2.4 = 1.5. The ratios are unequal, so EF is not parallel to QR.
(ii) PE/EQ = 4/4.5 = 8/9 and PF/FR = 8/9. The ratios are equal, so EF || QR by the converse of BPT.
(iii) Here whole sides are given, so use part-to-whole ratios. PE/PQ = 0.18/1.28 = 18/128 = 9/64 and PF/PR = 0.36/2.56 = 36/256 = 9/64. Equal, so EF || QR.
Example 3: A line parallel to the sides of a trapezium (NCERT Example 2)
ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB. Show that AE/ED = BF/FC.
Solution: Join AC, meeting EF at G.
AB || DC and EF || AB (given), so EF || DC (lines parallel to the same line are parallel to each other).
In ΔADC, EG || DC, so AE/ED = AG/GC (Theorem 6.1) … (1)
In ΔCAB, GF || AB, so CG/AG = CF/BF (Theorem 6.1). Inverting, AG/GC = BF/FC … (2)
From (1) and (2), AE/ED = BF/FC.
Example 4: Converse of BPT gives an isosceles triangle (NCERT Example 3)
S and T are points on sides PQ and PR of ΔPQR such that PS/SQ = PT/TR and ∠PST = ∠PRQ. Prove that PQR is an isosceles triangle.
Solution: PS/SQ = PT/TR (given), so ST || QR (Theorem 6.2).
Therefore ∠PST = ∠PQR (corresponding angles) … (1)
Also ∠PST = ∠PRQ (given) … (2)
From (1) and (2), ∠PRQ = ∠PQR. So PQ = PR (sides opposite equal angles), and ΔPQR is isosceles.
Example 5: Diagonals of a trapezium (NCERT Exercise 6.2, Q9)
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
Solution: Through O draw OE || DC, meeting AD at E. Then OE || AB as well.
In ΔADC, OE || DC, so AE/ED = AO/OC (BPT) … (1)
In ΔDAB, OE || AB, so DE/EA = DO/OB (BPT). Inverting, AE/ED = BO/OD … (2)
From (1) and (2), AO/OC = BO/OD. Rearranging, AO/BO = CO/DO.
Example 6: Parallel lines and crossing segments (NCERT Example 4)
Segments PS and QR intersect at O, and PQ || RS. Prove that ΔPOQ ~ ΔSOR.
Solution: PQ || RS (given).
So ∠P = ∠S (alternate angles, transversal PS) and ∠Q = ∠R (alternate angles, transversal QR).
Also ∠POQ = ∠SOR (vertically opposite angles).
Therefore ΔPOQ ~ ΔSOR (AAA similarity criterion). Note the order: P ↔ S, O ↔ O, Q ↔ R.
Example 7: A product of lengths gives SAS (NCERT Example 6)
Segments AC and BD intersect at O, and OA · OB = OC · OD. Show that ∠A = ∠C and ∠B = ∠D.
Solution: OA · OB = OC · OD (given). Dividing both sides by OC · OB: OA/OC = OD/OB … (1)
∠AOD = ∠COB (vertically opposite angles) … (2)
In ΔAOD and ΔCOB, the sides OA, OD include ∠AOD and the sides OC, OB include ∠COB. From (1) and (2), ΔAOD ~ ΔCOB (SAS similarity criterion).
So ∠A = ∠C and ∠D = ∠B (corresponding angles of similar triangles).
The proof is unchanged if the crossing segments are AB and CD instead.
Example 8: Angles from a similarity statement (NCERT Exercise 6.3, Q2)
Segments AC and DB intersect at O. ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Solution: D, O and B lie on one line, so ∠DOC and ∠BOC form a linear pair. ∠DOC = 180° − 125° = 55°.
In ΔODC, ∠DCO = 180° − ∠CDO − ∠DOC = 180° − 70° − 55° = 55°.
ΔODC ~ ΔOBA gives O ↔ O, D ↔ B, C ↔ A. So ∠OAB = ∠OCD = 55°.
Example 9: The girl and the lamp-post (NCERT Example 7)
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Solution: Let AB be the lamp-post (A the lamp, B its foot) and CD the girl after 4 seconds (C her head, D her feet). Let the shadow DE be x m, where E is the tip of the shadow.
BD = 1.2 m/s × 4 s = 4.8 m. Girl’s height CD = 90 cm = 0.9 m.
In ΔABE and ΔCDE: ∠B = ∠D (each 90°, both stand vertical to the ground) and ∠E = ∠E (same angle). So ΔABE ~ ΔCDE (AA).
Therefore BE/DE = AB/CD, that is (4.8 + x)/x = 3.6/0.9 = 4.
4.8 + x = 4x, so 3x = 4.8 and x = 1.6.
The girl’s shadow after 4 seconds is 1.6 m long.
Example 10: Height of a tower from shadows (NCERT Exercise 6.3, Q15)
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution: Let the tower have height h m. Both stand at 90° to the ground and the sun’s rays make equal angles with the ground at the same time, so the two triangles are similar by AA.
height of pole / shadow of pole = height of tower / shadow of tower
6/4 = h/28, so h = 28 × 6/4 = 28 × 1.5 = 42 m.
Example 11: Proving CA2 = CB · CD (NCERT Exercise 6.3, Q13)
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.
Solution: In ΔADC and ΔBAC:
∠ADC = ∠BAC (given) and ∠ACD = ∠BCA (common angle at C).
So ΔADC ~ ΔBAC (AA). The correspondence is A ↔ B, D ↔ A, C ↔ C.
Corresponding sides are proportional: AD/BA = DC/AC = AC/BC.
From DC/AC = AC/BC, cross-multiplying gives AC2 = BC × DC, that is CA2 = CB · CD.
Competency-Based Questions (with answers)
1. Case-based: Measuring a school flagpole
Class 10 students want the height of the school flagpole without climbing it. At 10 a.m. a student 1.5 m tall stands on level ground and her shadow measures 1.2 m. At the same moment the flagpole’s shadow measures 9.6 m.
(a) Why are the two triangles similar? The student and the flagpole both stand at 90° to the ground, and at the same moment the sun’s rays make equal angles with the ground. Two pairs of equal angles, so the triangles are similar by AA.
(b) Find the height of the flagpole. height/shadow is the same for both: 1.5/1.2 = h/9.6, so h = 9.6 × 1.5/1.2 = 9.6 × 1.25 = 12 m.
(c) Later in the day the student’s shadow is 3 m long. How long is the flagpole’s shadow then? The ratio height/shadow for the student is now 1.5/3 = 1/2, so the flagpole’s shadow is 12 × 2 = 24 m.
2. Case-based: Enlarging a photograph
A photographer prints a picture from 35 mm film and enlarges it to 45 mm. In the smaller print, the width of a door is 14 mm and the angle between the door frame and the floor line is 90°.
(a) What is the scale factor from the smaller print to the bigger print? 45/35 = 9/7.
(b) What is the width of the door in the bigger print? 14 × 9/7 = 18 mm.
(c) What is the angle between the door frame and the floor line in the bigger print? Give a reason. 90°. Similar figures have equal corresponding angles.
3. Assertion-Reason
Assertion (A): A square of side 4 cm and a rhombus of side 6 cm with an angle of 60° are similar.
Reason (R): Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are in the same ratio.
Choose: (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 false but R is true. The sides are in one ratio (4/6 each), but the square’s angles are 90° and the rhombus has a 60° angle. R is the correct definition, and it shows why A fails.
4. Assertion-Reason
Assertion (A): In ΔABC, D and E lie on AB and AC with AD = 3 cm, DB = 4 cm, AE = 4.5 cm and EC = 6 cm. Then DE || BC.
Reason (R): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Choose: (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: (a) AD/DB = 3/4 and AE/EC = 4.5/6 = 3/4. The ratios are equal, and R (Theorem 6.2) is exactly the reason DE || BC.
5. Assertion-Reason
Assertion (A): If two angles of one triangle are equal to two angles of another triangle, the triangles are similar.
Reason (R): Two triangles are similar only when all three pairs of corresponding sides have been measured and found equal.
Choose: (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: (c) A is true: it is the AA criterion, and the third angles are equal by the angle sum property. R is false: similar triangles need corresponding sides in the same ratio. Equal corresponding sides describe congruent triangles.
6. Source-based: Reading Activity 3
A student draws an angle XAY, marks five equal steps on each arm (B1 to B on AX and C1 to C on AY) and joins B3C3 and BC.
(a) What is the ratio AB3/B3B? AB3 is 3 steps and B3B is 2 steps, so the ratio is 3/2. AC3/C3C is also 3/2.
(b) What can you conclude about B3C3 and BC, and which theorem justifies it? B3C3 || BC, by the converse of BPT (Theorem 6.2), since the line divides AB and AC in the same ratio.
(c) If AB = 10 cm, find B3C3 : BC. B3C3 || BC, so ΔAB3C3 ~ ΔABC (AA, corresponding angles and the common angle A). So B3C3/BC = AB3/AB = 6/10 = 3 : 5.
7. Error analysis: The wrong correspondence
In ΔABC, ∠A = 40°, ∠B = 60°. In ΔPQR, ∠P = 60°, ∠Q = 80°. A student writes “ΔABC ~ ΔPQR, so AB/PQ = BC/QR.” Find the error and write the correct statement.
Answer: ∠C = 180° − 40° − 60° = 80° and ∠R = 180° − 60° − 80° = 40°. So ∠A = ∠R, ∠B = ∠P, ∠C = ∠Q. The triangles are similar by AA, but the correct statement is ΔABC ~ ΔRPQ, giving AB/RP = BC/PQ = CA/QR. The student matched letters by position instead of by equal angles.
8. Case-based: A ramp support
A ramp AC rises from point A on the ground to the top C of a vertical wall BC. A vertical support DE is fixed under the ramp, with D on the ground AB and E on the ramp AC. AD = 2 m, DE = 0.5 m and AB = 6 m.
(a) Why is DE || BC? Both are vertical, so both are perpendicular to the ground AB. Lines perpendicular to the same line are parallel.
(b) Name the similar triangles and the criterion. ΔADE ~ ΔABC by AA: ∠A is common and ∠ADE = ∠ABC = 90°.
(c) Find the height of the wall BC. DE/BC = AD/AB, so 0.5/BC = 2/6 and BC = 0.5 × 6/2 = 1.5 m.
Important Questions for Board Exams
1-Mark Questions
- Fill in the blanks (NCERT Exercise 6.1, Q1): (i) All circles are ____. (congruent, similar) (ii) All squares are ____. (similar, congruent) (iii) All ____ triangles are similar. (isosceles, equilateral) (i) similar (ii) similar (iii) equilateral.
- Is every pair of similar figures congruent? No. Similar figures have the same shape, but their sizes may differ. Every pair of congruent figures is similar.
- In ΔABC, DE || BC with D on AB and E on AC. If AD/DB = 3/5 and AE = 4.5 cm, find EC. By BPT, AE/EC = 3/5, so EC = 4.5 × 5/3 = 7.5 cm.
- If ΔABC ~ ΔDEF, AB = 4 cm, DE = 6 cm and EF = 9 cm, find BC. AB/DE = BC/EF, so 4/6 = BC/9 and BC = 6 cm.
- If ΔABC ~ ΔPQR, ∠A = 50° and ∠B = 60°, find ∠R. ∠R = ∠C = 180° − 50° − 60° = 70°.
2-Mark Questions
- S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS. (NCERT Exercise 6.3, Q5) In ΔRPQ and ΔRTS, ∠RPQ = ∠RTS (given) and ∠PRQ = ∠TRS (common angle R). So ΔRPQ ~ ΔRTS by AA.
- Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (NCERT Exercise 6.2, Q7) In ΔABC, let D be the mid-point of AB and DE || BC with E on AC. By BPT, AD/DB = AE/EC. AD = DB, so AD/DB = 1, hence AE/EC = 1 and AE = EC. So DE bisects AC.
- Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (NCERT Exercise 6.2, Q8) Let D and E be the mid-points of AB and AC. Then AD = DB and AE = EC, so AD/DB = 1 = AE/EC. The line DE divides AB and AC in the same ratio, so DE || BC by Theorem 6.2.
- Are the two triangles with sides 4 cm, 6 cm, 7 cm and 8 cm, 12 cm, 14 cm similar? Give a reason. Pair the sides in increasing order: 4/8 = 6/12 = 7/14 = 1/2. All three ratios are equal, so the triangles are similar by the SSS criterion.
3-Mark Questions
- The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium. (NCERT Exercise 6.2, Q10) From AO/BO = CO/DO we get AO/CO = BO/DO … (1). Through O draw OE || AB, meeting AD at E. In ΔDAB, OE || AB, so by BPT DE/EA = DO/OB, that is AE/ED = BO/OD … (2). From (1) and (2), AE/ED = AO/OC. So in ΔADC, the line EO divides AD and AC in the same ratio, and EO || DC by the converse of BPT. Now AB || EO and EO || DC, so AB || DC and ABCD is a trapezium.
- In ΔABC, AD and CE are altitudes, with D on BC (AD ⊥ BC) and E on AB (CE ⊥ AB). The altitudes AD and CE intersect each other at the point P inside the triangle. Show that: (i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (NCERT Exercise 6.3, Q7, parts i and ii) (i) In ΔAEP and ΔCDP, ∠AEP = ∠CDP = 90° and ∠APE = ∠CPD (vertically opposite). So ΔAEP ~ ΔCDP (AA). (ii) In ΔABD and ΔCBE, ∠ADB = ∠CEB = 90° and ∠ABD = ∠CBE (common angle B). So ΔABD ~ ΔCBE (AA).
- E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF. (NCERT Exercise 6.3, Q11) AB = AC, so ∠ABC = ∠ACB (angles opposite equal sides). E lies on CB produced, so ∠ECF is the same angle as ∠ACB, and ∠ABD is the same angle as ∠ABC. Hence ∠ABD = ∠ECF. Also ∠ADB = ∠EFC = 90°. So ΔABD ~ ΔECF (AA).
5-Mark Questions
- State and prove the Basic Proportionality Theorem. Statement: if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Given ΔABC with DE || BC, D on AB, E on AC. To prove AD/DB = AE/EC. Construction: join BE and CD, draw DM ⊥ AC and EN ⊥ AB. Proof: ar(ADE) = ½ AD × EN and ar(BDE) = ½ DB × EN, so ar(ADE)/ar(BDE) = AD/DB. Also ar(ADE) = ½ AE × DM and ar(DEC) = ½ EC × DM, so ar(ADE)/ar(DEC) = AE/EC. ΔBDE and ΔDEC are on the same base DE and between the same parallels BC and DE, so ar(BDE) = ar(DEC). Hence AD/DB = AE/EC.
- If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that AB/PQ = AD/PM. (NCERT Exercise 6.3, Q16) ΔABC ~ ΔPQR gives ∠B = ∠Q and AB/PQ = BC/QR. D and M are mid-points of BC and QR, so BC = 2BD and QR = 2QM. Then AB/PQ = 2BD/2QM = BD/QM. In ΔABD and ΔPQM, ∠ABD = ∠PQM and the sides including these angles are proportional (AB/PQ = BD/QM). So ΔABD ~ ΔPQM by SAS, and hence AB/PQ = AD/PM.
Common Mistakes and Examiner Tips
- Writing the vertices in the wrong order. ΔABC ~ ΔDEF means A ↔ D, B ↔ E, C ↔ F. Before writing the statement, find which angle equals which, then write the letters in that order.
- Mixing part and whole in BPT. AD/DB = AE/AC is wrong. Use part/part on both sides (AD/DB = AE/EC) or part/whole on both sides (AD/AB = AE/AC).
- Checking only one condition for quadrilaterals. A square and a rectangle have equal angles; a square and a rhombus have sides in one ratio. Neither pair is similar. For polygons with four or more sides, check angles and sides.
- Using SAS with an angle that is not included. The equal angle must lie between the two proportional sides. If it lies elsewhere, SAS does not apply.
- Testing parallel lines with decimals that are rounded. In Exercise 6.2 Q2 type questions, reduce each ratio to a fraction (such as 8/9 and 9/64) so that equality is exact.
- Taking the wrong base in lamp-post problems. The big triangle’s base is the distance from the foot of the lamp-post to the tip of the shadow (distance walked + shadow).
- Mixing units. In NCERT Example 7 the girl’s height is 90 cm and the lamp is 3.6 m high. Convert to one unit (0.9 m) before forming the ratio.
- Proof steps without reasons. Each line in a proof needs its reason in brackets: (given), (BPT), (alternate angles), (vertically opposite angles), (common angle), (AA similarity).
- Skipping the construction in the BPT proof. Write the construction line (join BE and CD, draw DM ⊥ AC and EN ⊥ AB) before using the areas. The heights EN and DM come from this step.
- Reading shadows at different times. Shadow ratios work only when both shadows are measured at the same time, because only then do the sun’s rays make equal angles with the ground.
Quick Revision Points
- Similar figures have the same shape but not necessarily the same size.
- All congruent figures are similar; similar figures need not be congruent.
- All circles, all squares and all equilateral triangles are similar.
- Two polygons with the same number of sides are similar if corresponding angles are equal and corresponding sides are in the same ratio.
- For polygons with four or more sides, either condition alone is not sufficient (square and rectangle, square and rhombus).
- ΔABC ~ ΔDEF: the order of letters gives the correspondence A ↔ D, B ↔ E, C ↔ F.
- Triangles with equal corresponding angles are called equiangular triangles.
- BPT (Theorem 6.1): DE || BC ⇒ AD/DB = AE/EC. Also called the Thales Theorem.
- Converse of BPT (Theorem 6.2): AD/DB = AE/EC ⇒ DE || BC.
- The mid-point theorem is the special case where both ratios equal 1.
- AAA (Theorem 6.3): equal corresponding angles ⇒ proportional sides ⇒ similar.
- AA: two pairs of equal angles are enough, because the third pair then matches by the angle sum property.
- SSS (Theorem 6.4): all three pairs of sides in the same ratio ⇒ similar.
- SAS (Theorem 6.5): one equal angle with the sides including it proportional ⇒ similar.
- Diagonals of a trapezium ABCD (AB || DC) meeting at O: AO/BO = CO/DO.
- D on BC with ∠ADC = ∠BAC ⇒ CA2 = CB · CD.
- Shadows at the same time: height/shadow is the same for every vertical object.
Weightage in Board Exams
Triangles is part of the geometry portion of the Class 10 Maths syllabus. Check the current CBSE course structure for the marks allotted to each unit. The 2026-27 NCERT chapter covers similar figures, BPT and its converse, and the AAA, AA, SSS and SAS criteria, with RHS as a note to the reader.
| Question type | What is usually asked from this chapter |
|---|---|
| MCQ and fill in the blank | Which figures are always similar; a missing length from BPT; a missing side or angle from a similarity statement |
| Assertion-Reason | Converse of BPT with given lengths; whether both conditions are needed for polygons; the AA criterion |
| Short answer | Testing EF || QR from four lengths; mid-point results from BPT; an AA proof with a common or vertically opposite angle |
| Long answer | Statement and proof of BPT; proofs with medians or bisectors of similar triangles; trapezium and quadrilateral results |
| Case-based | Shadows, lamp-posts, ramps and enlarged photographs: identify the similar triangles, name the criterion, then calculate |
Two skills carry most answers in this chapter: writing a proof with a reason on every line, and setting up the correct ratio from the correct correspondence. Work through every NCERT example and exercise question first.
Class 10 Maths Β· Chapter 6 β swipe through all 10 cards to understand the whole chapter.
Similar Figures
Figures with the same shape but not necessarily the same size.
All congruent figures are similar; similar figures need not be congruent.
- All circles, all squares and all equilateral triangles are similar.
- A circle and a square can never be similar.
- Photos of one monument in stamp, passport and postcard size are similar.
Similar Polygons
Same number of sides, corresponding angles equal and corresponding sides in the same ratio.
The common ratio is the scale factor, or Representative Fraction.
- Square and rectangle: equal angles, sides not in one ratio, not similar.
- Square and rhombus: sides in one ratio, angles unequal, not similar.
- Maps and building blueprints are drawn to a scale factor.
Writing ΞABC ~ ΞDEF
The order of letters tells you which vertex matches which.
Triangles with equal corresponding angles are called equiangular.
- ΞBAC ~ ΞEDF says the same as ΞABC ~ ΞDEF.
- ΞABC ~ ΞEDF would be a different, wrong claim here.
- Find the equal angles first, then write the letters in that order.
Basic Proportionality Theorem
A line parallel to one side of a triangle divides the other two sides in the same ratio.
Also called the Thales Theorem. Part-to-whole form: AD/AB = AE/AC.
- Proof uses areas: ar(BDE) = ar(DEC), same base DE, same parallels.
- AD = 2.4, DB = 3.6, AE = 2 gives EC = 3 cm.
- Never mix part/part with part/whole in one equation.
Converse of BPT
If a line divides two sides of a triangle in the same ratio, it is parallel to the third side.
Use exact fractions: PE/EQ = 4/4.5 = 8/9 and PF/FR = 8/9, so EF β₯ QR.
- PE/EQ = 1.3 and PF/FR = 1.5: EF is not parallel to QR.
- Mid-points give ratio 1 on both sides, so the joining line is parallel.
- Proved by contradiction: E and Eβ² must coincide.
AAA and AA Criteria
Equal corresponding angles make the sides proportional and the triangles similar.
Two angles are enough: the third matches by the angle sum property.
- Look for common angles and vertically opposite angles.
- Parallel lines give alternate and corresponding angles.
- Altitudes, poles and walls give right angles.
SSS and SAS Criteria
All sides proportional (SSS), or one equal angle between two proportional sides (SAS).
For SAS the equal angle must be the included angle.
- Sides 3, 6, 8 and 4.5, 9, 12: every ratio 2/3, similar by SSS.
- OA Β· OB = OC Β· OD gives OA/OC = OD/OB for an SAS proof.
- RHS: right triangles with hypotenuse and one side proportional.
Heights from Shadows
At the same moment, height divided by shadow is the same for every vertical object.
Both shadows must be measured at the same time.
- Pole 6 m with shadow 4 m; tower shadow 28 m gives tower 42 m.
- The triangles are similar by AA: a right angle and the sun’s angle.
- Heights of mountains are found this way, by indirect measurement.
The Girl and the Lamp
The shadow tip is common to both triangles, so they are similar by AA.
The big triangle’s base is the distance walked plus the shadow.
- Girl 0.9 m, lamp 3.6 m, walks 1.2 m/s for 4 s, so d = 4.8 m.
- (4.8 + x)/x = 4 gives x = 1.6 m.
- Convert 90 cm to 0.9 m before forming the ratio.
Results to Remember
Standard NCERT results that come straight from the similarity criteria.
Give a reason in brackets on every line of a proof.
- Trapezium diagonals meeting at O: AO/BO = CO/DO.
- Similar triangles: medians are in the ratio of corresponding sides.
- E on AD produced of parallelogram ABCD, BE meets CD at F: ΞABE ~ ΞCFB.
π Practice Triangles - 10 board questions
CBSE previous-year and competency-based Β· with answers & explanations
Start βClose β
Why D: Step 1 (scale factor from the given sides): DE/AB = 2. Step 2 (corresponding sides in the same ratio): EF/BC = DE/AB = 2, so EF = 2 Γ 8 = 16 cm.
Why not A: 4 cm halves BC, but ΞDEF is the larger triangle.
Why not B: 8 cm makes EF equal to BC, which would need the triangles to be congruent.
Why not C: 12 cm adds 4 cm to BC instead of doubling it.
Remember: Similar means multiply by the scale factor, never add a fixed amount.
Why A: Step 1 (BPT): AE/EC = AD/DB = 1/3. Step 2 (part-to-whole): AE is 1 part out of 1 + 3 = 4, so AE = 6/4 = 1.5 cm.
Why not B: 1 cm leaves EC = 5 cm, a ratio of 1/5, not 1/3.
Why not C: 2 cm is 6 Γ 1/3, treating 1/3 as AE/AC instead of AE/EC.
Why not D: 3 cm is half of AC, true only if D were the mid-point of AB.
Remember: 1 : 3 means 4 equal parts in all.
Why C: in ΞABC ~ ΞEDF the correspondence is AβE, BβD, CβF. So β A = β E, not β D. β C = β F is correct, but the pair β A = β D makes the statement false.
Why not A: perimeters of similar triangles are in the ratio of corresponding sides, so A is true.
Why not B: ABβED and ACβEF are corresponding pairs, so B is true.
Why not D: with AB = kΒ·ED, BC = kΒ·DF and AC = kΒ·EF, the k cancels in (AB + BC)/AC, so D is true.
Remember: Letters in the same position correspond: 1stβ1st, 2ndβ2nd, 3rdβ3rd.
Why D: the sides including β C are CA and CB, and those including β R are RP and RQ. SAS needs AC/PR = BC/QR, which rearranges to AC/BC = PR/QR.
Why not A: AC and AB include β A, not β C.
Why not B: BC/AC = PR/QR rearranges to BC/PR = AC/QR, pairing BC with PR, which do not correspond.
Why not C: PQ does not touch R, so it is not a side including β R.
Remember: SAS: the angle sits between the two proportional sides.
Why A: Step 1 (AA similarity, corresponding angles from PQ β₯ YZ): ΞXPQ ~ ΞXYZ. Step 2 (part-to-whole): XP/XY = 2/(2 + 3) = 2/5. Step 3 (corresponding sides in the same ratio): PQ/YZ = 2/5, so YZ = 5 Γ 5/2 = 12.5 cm.
Why not B: 10 cm just multiplies PQ by 2, reading one number of the ratio as the multiplier.
Why not C: 15 cm multiplies PQ by 3.
Why not D: 7.5 cm uses 5 Γ 3/2, the part-to-part ratio PY/XP, instead of XY/XP.
Remember: Compare the parallel segments with the part-to-whole ratio XP/XY.
Why B: a rhombus need not have right angles, for example one with angles 60Β° and 120Β°. Its sides are in one ratio with the square’s sides, but its angles differ, so it is not similar to the square. Equal area does not change this.
Why not A: this is true, congruent figures have the same shape and size; its converse “similar figures are congruent” is the false one.
Why not C: this is true, every equilateral triangle has angles 60Β°, 60Β°, 60Β°.
Why not D: this is true, similar triangles can differ in size.
Remember: For polygons, equal angles AND proportional sides are both needed.
Why D: Step 1 (BPT): AQ/QC = AP/PB = 4/13. Step 2 (part-to-whole): AQ is 4 parts out of 4 + 13 = 17, so AQ = (4/17) Γ 20.4 = 4.8 cm. Check: QC = 15.6 cm and 4.8/15.6 = 4/13.
Why not A: 2.8 cm leaves QC = 17.6 cm, and 2.8/17.6 is not 4/13.
Why not B: 5.8 cm leaves QC = 14.6 cm, and 5.8/14.6 is not 4/13.
Why not C: 3.8 cm leaves QC = 16.6 cm, and 3.8/16.6 is not 4/13.
Remember: A ratio a : b splits the whole into a + b parts.
Why D: let the diagonals of ABCD meet at O with AO/OC = BO/OD. In ΞAOB and ΞCOD, β AOB = β COD (vertically opposite) and the sides including these angles are proportional, so the triangles are similar (SAS). Then β OAB = β OCD, which are alternate angles, so AB β₯ DC. A quadrilateral with a pair of parallel sides is a trapezium.
Why not A: a parallelogram needs the diagonals to bisect each other (ratio 1 : 1); a ratio such as 2 : 3 still satisfies the condition but gives only one pair of parallel sides.
Why not B: a rectangle also needs equal diagonals and right angles, which the condition does not give.
Why not C: a square needs equal sides and right angles too, none of which follows.
Remember: Proportional diagonals give a trapezium; equal halves give a parallelogram.
Why B: Step 1 (similar triangles, perimeter ratio = side ratio): PQ/AB = perimeter of ΞPQR / perimeter of ΞABC = 48/56. Step 2 (simplify, dividing by 8): 48/56 = 6/7.
Why not A: 7/8 does not come from 48 and 56 at all; dividing both by 8 gives 6 and 7.
Why not C: 7/6 is AB/PQ, the ratio turned upside down.
Why not D: 8/7 also mixes the divisor 8 into the answer.
Remember: Keep the same triangle on top in both ratios: PQR over ABC.
Why B: AB and BC meet at B, so their included angle is β B. DE and FD meet at D, so their included angle is β D. With β B = β D and AB/DE = BC/FD, SAS gives ΞABC ~ ΞEDF.
Why not A: β A is not between AB and BC, so it is not the included angle.
Why not C: β E is not between DE and FD; matching B with E just follows the letter positions.
Why not D: neither β A nor β F is included between the proportional sides.
Remember: For SAS, find the letter common to both proportional sides: that vertex is the angle.
Chapter Navigation
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Related Chapters in Class 10 Maths
- Real Numbers Class 10 Notes
- Polynomials Class 10 Notes
- Pair of Linear Equations in Two Variables Class 10 Notes
- Quadratic Equations Class 10 Notes
- Arithmetic Progressions Class 10 Notes
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Frequently Asked Questions
Congruent figures have the same shape and the same size. Similar figures have the same shape but not necessarily the same size. So every pair of congruent figures is similar, but similar figures need not be congruent. For example, all circles are similar, while only circles with equal radii are congruent.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. In triangle ABC with DE parallel to BC, AD/DB = AE/EC. It is also called the Thales Theorem, and it is proved using areas of triangles on the same base between the same parallels.
Use the converse of BPT. Work out the ratio in which the line divides each of the two sides, as exact fractions. If AD/DB = AE/EC, then DE is parallel to BC. If the ratios are unequal, the line is not parallel, because a parallel line would have to give equal ratios by BPT.
AAA: corresponding angles equal. AA: two pairs of angles equal, since the third pair then matches by the angle sum property. SSS: all three pairs of corresponding sides in the same ratio. SAS: one angle equal and the sides including that angle in the same ratio. The NCERT note to the reader adds RHS for right triangles.
The order shows which vertices correspond. Writing triangle ABC ~ triangle DEF means angle A equals angle D, B equals E and C equals F, and AB/DE = BC/EF = CA/FD. If angle A actually equals angle E, the statement must be rewritten, or every ratio you read from it will be wrong.
A square and a rectangle have equal angles but sides not in one ratio, and a square and a rhombus have sides in one ratio but unequal angles, so neither pair is similar. For triangles, Theorems 6.3 and 6.4 show that equal angles force proportional sides and proportional sides force equal angles, so one condition is enough.
At the same moment the sun’s rays make equal angles with the ground, and a pole and a tower both stand at 90 degrees. The two triangles are similar by AA, so height divided by shadow is the same for both. A 6 m pole with a 4 m shadow and a tower with a 28 m shadow give a tower height of 42 m.
