Quadrilaterals (listed in the syllabus as 4-gons) starts with a precise definition of a 4-gon and its 360° angle sum, then proves the properties of a parallelogram and the four tests that show a 4-gon is one. It covers rectangles, rhombuses and squares, the symmetries of 4-gons and why any 4-gon tiles a plane. The midpoint theorem and its converse lead to two striking results: the midpoints of any 4-gon form a parallelogram, and the medians of a triangle meet at a point dividing each in the ratio 2 : 1.
Key Concepts
1. What a 4-gon Is and Why Its Angles Add to 360°
Take four points A, B, C and D in a plane, with no three of them on one straight line. Join them in order by the segments AB, BC, CD and DA. If these four segments meet only at their shared end points, so that AB does not cross CD and BC does not cross DA, the closed figure ABCD is a 4-gon, also called a quadrilateral.
| Part of 4-gon ABCD | What it is |
|---|---|
| Vertices | A, B, C, D |
| Sides | AB, BC, CD, DA |
| Angles | ∠A (∠DAB), ∠B (∠ABC), ∠C (∠BCD), ∠D (∠CDA) |
| Diagonals | AC and BD, joining opposite vertices |
| Adjacent sides | Sides with a common vertex, such as AB and BC |
| Opposite sides | Sides with no common vertex: AB and CD, BC and DA |
| Opposite angles | ∠A and ∠C, ∠B and ∠D |
The name ABCD gives the order of the vertices round the figure, so ABCD and ABDC are different 4-gons.
A 4-gon is convex when each angle is less than 180° and both diagonals lie inside it. A concave 4-gon has one reflex angle.
Angle sum property. The angles of a 4-gon add up to 360°.
Proof. Choose a diagonal that lies inside the 4-gon, say AC. It splits ABCD into △ABC and △ACD. At vertex A, ∠DAB = ∠BAC + ∠CAD, and at C, ∠BCD = ∠BCA + ∠ACD. So the six angles of the two triangles make up exactly the four angles of the 4-gon. Each triangle has angle sum 180°, so
∠A + ∠B + ∠C + ∠D = 180° + 180° = 360°
Consequences:
- A 4-gon cannot have all four angles acute, because four angles each less than 90° add to less than 360°.
- It cannot have four obtuse angles either, since they would add to more than 360°. At most three angles can be obtuse.
- If three angles are known, the fourth is 360° minus their sum.
A 4-gon is not rigid. A frame of four hinged sticks can change shape while its sides stay the same, unlike a triangle. A diagonal brace splits it into two rigid triangles, which is why gates use one.
2. The Parallelogram and Its Properties
A parallelogram is a 4-gon in which both pairs of opposite sides are parallel: AB ∥ DC and AD ∥ BC.
Every property of a parallelogram comes from one theorem.
Theorem 1. A diagonal divides a parallelogram into two congruent triangles.
Proof. In parallelogram ABCD, draw diagonal AC.
- AB ∥ DC and AC is a transversal, so ∠BAC = ∠DCA (alternate angles).
- AD ∥ BC and AC is a transversal, so ∠BCA = ∠DAC (alternate angles).
- AC = CA (common side).
So △ABC ≅ △CDA by the ASA rule. The correspondence is A ↔ C, B ↔ D, C ↔ A.
Theorem 2. In a parallelogram, opposite sides are equal. From △ABC ≅ △CDA, by CPCT, AB = CD and BC = DA.
Theorem 3. In a parallelogram, opposite angles are equal. From the same congruence, ∠B = ∠D. Also ∠A = ∠BAC + ∠CAD = ∠DCA + ∠BCA = ∠C. (Using diagonal BD instead gives ∠A = ∠C directly.)
Adjacent angles are supplementary. AD ∥ BC with transversal AB, so ∠A and ∠B are co-interior angles and ∠A + ∠B = 180°. The same holds for every pair of adjacent angles. So if one angle of a parallelogram is 70°, the angles are 70°, 110°, 70° and 110°.
Theorem 4. The diagonals of a parallelogram bisect each other.
Proof. Let the diagonals AC and BD meet at O. In △AOB and △COD:
- AB = CD (opposite sides of a parallelogram)
- ∠OAB = ∠OCD (alternate angles, AB ∥ DC, transversal AC)
- ∠OBA = ∠ODC (alternate angles, AB ∥ DC, transversal BD)
So △AOB ≅ △COD (ASA), and by CPCT, OA = OC and OB = OD. O is the midpoint of both diagonals.
The diagonals of a general parallelogram are unequal and not perpendicular. Those extra facts belong to the rectangle and the rhombus.
3. Tests for a Parallelogram (Characterisations)
A converse of a theorem swaps its hypothesis and conclusion. “If ABCD is a parallelogram, then its opposite sides are equal” has the converse “If the opposite sides of a 4-gon are equal, then it is a parallelogram”. A converse must be proved separately, because some converses are false. For the parallelogram, four useful converses are true, and each one gives a test.
Test 1. If both pairs of opposite sides of a 4-gon are equal, it is a parallelogram.
Given AB = CD and AD = BC. Draw diagonal AC. In △ABC and △CDA: AB = CD, BC = DA, AC = CA. So △ABC ≅ △CDA (SSS). By CPCT, ∠BAC = ∠DCA; these are alternate angles for lines AB and DC with transversal AC, so AB ∥ DC. Also ∠BCA = ∠DAC, so BC ∥ AD. Both pairs of opposite sides are parallel.
Test 2. If both pairs of opposite angles of a 4-gon are equal, it is a parallelogram.
Given ∠A = ∠C and ∠B = ∠D. The angle sum gives ∠A + ∠B + ∠C + ∠D = 360°, so 2(∠A + ∠B) = 360° and ∠A + ∠B = 180°. These are co-interior angles for lines AD and BC with transversal AB, so AD ∥ BC. In the same way, ∠A + ∠D = 180° gives AB ∥ DC.
Test 3. If the diagonals of a 4-gon bisect each other, it is a parallelogram.
Given OA = OC and OB = OD, where O is the meeting point of the diagonals. In △AOB and △COD: OA = OC, OB = OD, and ∠AOB = ∠COD (vertically opposite angles). So △AOB ≅ △COD (SAS), and ∠OAB = ∠OCD. These are alternate angles, so AB ∥ DC. The same argument with △AOD and △COB gives AD ∥ BC.
Test 4. If one pair of opposite sides of a 4-gon is equal and parallel, it is a parallelogram.
Given AB = CD and AB ∥ CD. Draw diagonal AC. ∠BAC = ∠DCA (alternate angles), AB = CD and AC is common, so △ABC ≅ △CDA (SAS). By CPCT, ∠BCA = ∠DAC, which are alternate angles, so BC ∥ AD.
Test 4 is the quickest in many proofs, because you only need to look at one pair of sides.
Conditions that are NOT enough.
- One pair of opposite sides parallel only gives a trapezium.
- One pair of opposite sides equal and the other pair parallel is also not enough. An isosceles trapezium has AB ∥ DC and AD = BC, but AB ≠ DC.
- Equal diagonals do not make a parallelogram. An isosceles trapezium has equal diagonals.
Before trusting a converse, try to build a counter example. One counter example is enough to show it is false.
4. Rectangle, Rhombus, Square and Other 4-gons
Special parallelograms add one more condition. Because they are parallelograms, they keep every property from Concept 2.
Rectangle. A parallelogram with one right angle. Since adjacent angles add to 180° and opposite angles are equal, all four angles are then 90°.
- The diagonals of a rectangle are equal. In △DAB and △CBA: AD = BC, AB = BA and ∠DAB = ∠CBA = 90°. So the triangles are congruent (SAS) and BD = AC.
- Converse: a parallelogram with equal diagonals is a rectangle. If AC = BD, then △DAB ≅ △CBA (SSS: AD = BC, AB common, BD = AC). So ∠A = ∠B. With ∠A + ∠B = 180°, each is 90°.
Rhombus. A parallelogram with a pair of adjacent sides equal. Since opposite sides are equal, all four sides are then equal.
- The diagonals of a rhombus bisect each other at right angles. In △AOB and △AOD: AB = AD, OB = OD and AO is common. So the triangles are congruent (SSS) and ∠AOB = ∠AOD. They form a linear pair, so each is 90°.
- Each diagonal bisects the angles at its ends. △ABC ≅ △ADC (SSS), so ∠BAC = ∠DAC.
- Converse: a parallelogram whose diagonals are perpendicular is a rhombus. If ∠AOB = ∠AOD = 90°, then △AOB ≅ △AOD (SAS: OB = OD, right angles, AO common), so AB = AD.
- With diagonals d1 and d2, the four right triangles at O have legs d1/2 and d2/2, so by Pythagoras, side2 = (d1/2)2 + (d2/2)2.
Square. A rectangle with all sides equal, so it is both a rectangle and a rhombus. Its diagonals are equal, bisect each other at right angles and bisect the 90° angles into 45°.
Other 4-gons.
- Trapezium: exactly one pair of opposite sides parallel. The two angles along each non-parallel side are co-interior, so each such pair adds to 180°.
- Isosceles trapezium: a trapezium whose non-parallel sides are equal. Its base angles are equal and its diagonals are equal.
- Kite: two pairs of equal adjacent sides (AB = AD and CB = CD). Its diagonals are perpendicular, and diagonal AC bisects diagonal BD.
| Property | Parallelogram | Rectangle | Rhombus | Square |
|---|---|---|---|---|
| Opposite sides parallel and equal | Yes | Yes | Yes | Yes |
| All sides equal | No | No | Yes | Yes |
| All angles 90° | No | Yes | No | Yes |
| Diagonals bisect each other | Yes | Yes | Yes | Yes |
| Diagonals equal | No | Yes | No | Yes |
| Diagonals perpendicular | No | No | Yes | Yes |
| Diagonals bisect vertex angles | No | No | Yes | Yes |
“No” in this table means “not in general”.
Illustration. A rhombus has diagonals 16 cm and 30 cm. Half-diagonals are 8 cm and 15 cm, so side = √(64 + 225) = √289 = 17 cm and the perimeter is 68 cm.
5. Central Symmetry, Reflection and Rotation Symmetry, and Tiling
A figure has central symmetry about a point O when, for every point P of the figure, the point P′ with O as the midpoint of PP′ also lies on the figure. This is the same as saying that a rotation through 180° (a half-turn) about O maps the figure onto itself.
Every parallelogram has central symmetry about the meeting point O of its diagonals. Since OA = OC and OB = OD, the half-turn about O sends A to C, C to A, B to D and D to B. So it sends side AB to side CD and side BC to side DA, and the parallelogram lands on itself.
The converse also holds. If a half-turn about O maps a 4-gon onto itself, it must swap opposite vertices, so O is the midpoint of AC and of BD. The diagonals bisect each other, so by Test 3 the 4-gon is a parallelogram. In short: a 4-gon has central symmetry exactly when it is a parallelogram, and the centre is where the diagonals meet.
Reflection (line) symmetry means a line divides the figure into two mirror halves. Rotation symmetry means a turn through less than 360° about a centre maps the figure onto itself.
| 4-gon | Lines of symmetry | Where the lines are | Smallest rotation that fits |
|---|---|---|---|
| Parallelogram (not a rectangle or rhombus) | 0 | None | 180° |
| Rectangle (not a square) | 2 | Through the midpoints of opposite sides | 180° |
| Rhombus (not a square) | 2 | Along the diagonals | 180° |
| Square | 4 | Both diagonals and both midlines | 90° |
| Kite (not a rhombus) | 1 | The diagonal through the vertices where equal sides meet | None below 360° |
| Isosceles trapezium | 1 | Through the midpoints of the parallel sides | None below 360° |
A common trap: a diagonal of a rectangle is not a line of symmetry. Folding a non-square rectangle along a diagonal does not make the halves match.
Any 4-gon can tile a plane. Tiling means covering the plane with copies of one shape, with no gaps and no overlaps. Give 4-gon ABCD a half-turn about the midpoint of a side; the copy fits exactly along that side. Repeat across every side. At each meeting point, four copies bring together the four different angles A, B, C and D, which add to 360°, a full turn. So there is no gap and no overlap, even for a concave 4-gon.
6. The Midpoint Theorem and Its Converse
Midpoint theorem. The line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of it.
Proof. In △ABC, let E be the midpoint of AB and F the midpoint of AC. We show EF ∥ BC and EF = BC/2.
- Extend EF beyond F to a point D so that FD = EF. Join CD.
- In △AEF and △CDF: AF = CF (F is the midpoint of AC), EF = DF (construction), and ∠AFE = ∠CFD (vertically opposite). So △AEF ≅ △CDF (SAS).
- By CPCT, CD = AE, and AE = EB, so CD = EB.
- Also by CPCT, ∠EAF = ∠DCF. These are alternate angles for lines AB and CD with transversal AC, so CD ∥ AB, that is, CD ∥ EB.
- In 4-gon EBCD, one pair of opposite sides, EB and CD, is equal and parallel. By Test 4, EBCD is a parallelogram.
- So ED ∥ BC, which means EF ∥ BC, and ED = BC. Since EF = FD, EF = ED/2 = BC/2.
Converse of the midpoint theorem. The line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side.
Proof. In △ABC, E is the midpoint of AB, and the line through E parallel to BC meets AC at F. Let F′ be the midpoint of AC. By the midpoint theorem, EF′ ∥ BC. Through the point E there is only one line parallel to BC (Playfair’s axiom), so the lines EF and EF′ are the same line. This line meets AC at only one point, so F = F′, and F is the midpoint of AC.
Uses of the theorem.
- Joining the three midpoints D, E, F of the sides of △ABC splits it into four congruent triangles, each with sides equal to half of the sides of ABC. The middle triangle DEF has perimeter half that of △ABC.
- In a trapezium, the line through the midpoint of one non-parallel side, drawn parallel to the parallel sides, passes through the midpoint of the other non-parallel side.
- Through the midpoint of the hypotenuse of a right triangle, a line parallel to one leg meets the other leg at its midpoint and is perpendicular to it. This shows the midpoint of the hypotenuse is equally far from all three vertices (Example 11).
Illustration. In △PQR, PQ = 10 cm, QR = 14 cm and RP = 8 cm. X and Y are the midpoints of PQ and PR. Then XY ∥ QR and XY = 14/2 = 7 cm.
7. The Midpoint 4-gon and the Medians of a Triangle
Theorem: the 4-gon formed by joining the midpoints of the sides of any 4-gon, in order, is a parallelogram.
Proof. Let P, Q, R and S be the midpoints of AB, BC, CD and DA of 4-gon ABCD. Draw diagonal AC.
- In △ABC, P and Q are midpoints of AB and BC, so PQ ∥ AC and PQ = AC/2 (midpoint theorem).
- In △ADC, S and R are midpoints of AD and DC, so SR ∥ AC and SR = AC/2.
- So PQ ∥ SR and PQ = SR. By Test 4, PQRS is a parallelogram.
Using diagonal BD in the same way gives PS = QR = BD/2. So the perimeter of PQRS = AC/2 + BD/2 + AC/2 + BD/2 = AC + BD, the sum of the diagonals of ABCD.
| Original 4-gon ABCD | Midpoint 4-gon PQRS | Reason |
|---|---|---|
| Any 4-gon | Parallelogram | PQ and SR both equal and parallel to half of AC |
| Diagonals perpendicular (rhombus, kite) | Rectangle | PQ ∥ AC and PS ∥ BD, so PQ ⊥ PS |
| Diagonals equal (rectangle, isosceles trapezium) | Rhombus | PQ = AC/2 = BD/2 = PS |
| Square | Square | Diagonals equal and perpendicular |
Medians of a triangle. A median joins a vertex of a triangle to the midpoint of the opposite side. Every triangle has three medians.
Theorem: the three medians of a triangle pass through one point, and this point divides each median in the ratio 2 : 1 from the vertex. The common point is the centroid, usually named G.
Proof. In △ABC, let E be the midpoint of AC and F the midpoint of AB. Let the medians BE and CF meet at G. Let P be the midpoint of BG and Q the midpoint of CG.
- In △ABC, F and E are midpoints of AB and AC, so FE ∥ BC and FE = BC/2.
- In △GBC, P and Q are midpoints of GB and GC, so PQ ∥ BC and PQ = BC/2.
- So FE ∥ PQ and FE = PQ. By Test 4, FEQP is a parallelogram.
- Its diagonals are FQ and EP. F, G, Q lie on median CF, and E, G, P lie on median BE, so the diagonals meet at G. The diagonals of a parallelogram bisect each other, so GE = GP and GF = GQ.
- P is the midpoint of BG, so BP = PG = GE. So BG = 2 GE, that is, BG : GE = 2 : 1. In the same way CQ = QG = GF, so CG : GF = 2 : 1.
- Apply the same argument to medians BE and AD, where D is the midpoint of BC. They meet at a point that divides BE in the ratio 2 : 1 from B. There is only one such point on BE, and it is G. So AD also passes through G, and AG : GD = 2 : 1.
So the three medians are concurrent at G, with AG = (2/3)AD and GD = (1/3)AD for each median.
Illustration. Median AD of △ABC is 15 cm. Then AG = (2/3) × 15 = 10 cm and GD = 5 cm.
8. Midpoints and Parallelograms with Coordinates
The midpoint of the segment joining A(x1, y1) and B(x2, y2) is
M = ((x1 + x2)/2, (y1 + y2)/2)
The midpoint of (2, 7) and (8, −3) is ((2 + 8)/2, (7 + (−3))/2) = (5, 2).
Finding the fourth vertex of a parallelogram. In parallelogram ABCD the diagonals AC and BD bisect each other, so they have the same midpoint:
(xA + xC)/2 = (xB + xD)/2 and (yA + yC)/2 = (yB + yD)/2
So xD = xA + xC − xB and yD = yA + yC − yB. In words: the sum of the coordinates of opposite vertices is the same for both pairs.
The order of the letters matters: in ABCD, D is opposite B. Three points given with no order allow three different parallelograms.
Testing a 4-gon with coordinates.
- Parallelogram: the midpoints of AC and BD are the same point (Test 3).
- Rectangle: a parallelogram whose diagonals are equal, checked with the distance formula √((x2 − x1)2 + (y2 − y1)2).
- Rhombus: a parallelogram with a pair of equal adjacent sides.
- Square: both of the above.
Formula and Theorem Sheet
| Result | Statement |
|---|---|
| Angle sum of a 4-gon | ∠A + ∠B + ∠C + ∠D = 360° |
| Diagonal of a parallelogram | △ABC ≅ △CDA |
| Parallelogram sides and angles | AB = CD, BC = DA; ∠A = ∠C, ∠B = ∠D; ∠A + ∠B = 180° |
| Parallelogram diagonals | OA = OC, OB = OD |
| Test 1 | Both pairs of opposite sides equal ⇒ parallelogram |
| Test 2 | Both pairs of opposite angles equal ⇒ parallelogram |
| Test 3 | Diagonals bisect each other ⇒ parallelogram |
| Test 4 | One pair of opposite sides equal and parallel ⇒ parallelogram |
| Rectangle | Diagonals equal; parallelogram with equal diagonals is a rectangle |
| Rhombus | Diagonals perpendicular bisectors of each other; side2 = (d1/2)2 + (d2/2)2 |
| Central symmetry | A 4-gon is centrally symmetric exactly when it is a parallelogram |
| Midpoint theorem | EF ∥ BC, EF = BC/2 |
| Converse of midpoint theorem | Line through midpoint of AB parallel to BC bisects AC |
| Midpoint 4-gon | PQRS is a parallelogram; perimeter = AC + BD |
| Centroid | Medians concurrent; AG : GD = 2 : 1 |
| Midpoint formula | ((x1 + x2)/2, (y1 + y2)/2) |
| Fourth vertex | D = (xA + xC − xB, yA + yC − yB) |
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Important Definitions
| Term | Definition |
|---|---|
| 4-gon (quadrilateral) | A closed figure formed by joining four points, no three collinear, in order by four segments that meet only at their end points |
| Diagonal | A segment joining two opposite vertices of a 4-gon |
| Parallelogram | A 4-gon with both pairs of opposite sides parallel |
| Rectangle | A parallelogram with one angle (and so every angle) a right angle |
| Rhombus | A parallelogram with a pair of adjacent sides (and so all sides) equal |
| Square | A rectangle with all sides equal; it is also a rhombus |
| Trapezium | A 4-gon with exactly one pair of opposite sides parallel |
| Kite | A 4-gon with two pairs of equal adjacent sides |
| Median | A segment joining a vertex of a triangle to the midpoint of the opposite side |
| Centroid | The point where the three medians of a triangle meet |
| Central symmetry | A figure has central symmetry about O if a half-turn about O maps it onto itself |
Solved Examples
Example 1: Angles in a ratio
The angles of a 4-gon are in the ratio 3 : 5 : 9 : 13. Find all the angles.
Solution: Let the angles be 3x, 5x, 9x and 13x. By the angle sum property, 3x + 5x + 9x + 13x = 360°, so 30x = 360° and x = 12°.
The angles are 36°, 60°, 108° and 156°. Check: 36 + 60 + 108 + 156 = 360.
Example 2: Angles of a parallelogram
In a parallelogram, one angle is 24° less than twice its adjacent angle. Find all four angles.
Solution: Let one angle be x. Its adjacent angle is 2x − 24°. Adjacent angles of a parallelogram are supplementary, so x + 2x − 24° = 180°, 3x = 204° and x = 68°.
The adjacent angle is 2(68°) − 24° = 112°. Opposite angles are equal, so the angles are 68°, 112°, 68° and 112°.
Example 3: Sides from a perimeter
The perimeter of a parallelogram is 56 cm and two adjacent sides are in the ratio 3 : 4. Find the sides.
Solution: Let the adjacent sides be 3k and 4k. Opposite sides are equal, so the perimeter is 2(3k + 4k) = 14k. Then 14k = 56 and k = 4.
The sides are 12 cm, 16 cm, 12 cm and 16 cm.
Example 4: Diagonals that bisect each other
The diagonals of parallelogram ABCD meet at O. OA = 2x + 3, OC = 5x − 6, OB = y + 4 and OD = 3y − 2 (all in cm). Find the lengths of AC and BD.
Solution: The diagonals of a parallelogram bisect each other, so OA = OC and OB = OD.
2x + 3 = 5x − 6 gives 3x = 9, x = 3. So OA = 9 cm and AC = 2 × 9 = 18 cm.
y + 4 = 3y − 2 gives 2y = 6, y = 3. So OB = 7 cm and BD = 2 × 7 = 14 cm.
Example 5: Side of a rhombus from its diagonals
The diagonals of a rhombus are 24 cm and 10 cm. Find its side and perimeter.
Solution: The diagonals of a rhombus bisect each other at right angles, so they form four right triangles with legs 12 cm and 5 cm.
side2 = 122 + 52 = 144 + 25 = 169, so side = 13 cm. Perimeter = 4 × 13 = 52 cm.
Example 6: Bisectors of adjacent angles
In parallelogram ABCD, the bisectors of ∠A and ∠B meet at P. Prove that ∠APB = 90°.
Solution: AD ∥ BC, so ∠A + ∠B = 180° (co-interior angles). Halving, ∠PAB + ∠PBA = (1/2)∠A + (1/2)∠B = 90°.
In △APB, ∠APB = 180° − (∠PAB + ∠PBA) = 180° − 90° = 90°. Doing this at all four corners shows that, when the parallelogram is not a rhombus, its four angle bisectors enclose a 4-gon with four right angles, a rectangle.
Example 7: Perimeter of the midpoint triangle
The sides of △ABC are 8 cm, 10 cm and 12 cm. D, E and F are the midpoints of the sides. Find the perimeter of △DEF.
Solution: By the midpoint theorem, each side of △DEF is half of the side of △ABC parallel to it. So the sides of △DEF are 4 cm, 5 cm and 6 cm.
Perimeter = 4 + 5 + 6 = 15 cm, half the perimeter of △ABC (30 cm).
Example 8: Parts of a median
In △PQR, the median PM is 12 cm long and G is the centroid. Find PG and GM.
Solution: The centroid divides each median in the ratio 2 : 1 from the vertex. So PG = (2/3) × 12 = 8 cm and GM = (1/3) × 12 = 4 cm. Check: PG : GM = 8 : 4 = 2 : 1.
Example 9: Fourth vertex of a parallelogram
A(1, 2), B(4, 3) and C(6, 6) are three vertices of parallelogram ABCD. Find D.
Solution: The diagonals AC and BD have the same midpoint. Midpoint of AC = ((1 + 6)/2, (2 + 6)/2) = (3.5, 4).
Let D = (x, y). Midpoint of BD = ((4 + x)/2, (3 + y)/2). So (4 + x)/2 = 3.5, giving x = 3, and (3 + y)/2 = 4, giving y = 5.
D = (3, 5). Shortcut check: xD = 1 + 6 − 4 = 3 and yD = 2 + 6 − 3 = 5.
Example 10: Parallelogram or rectangle
Show that A(−2, −1), B(1, 0), C(4, 3) and D(1, 2) are the vertices of a parallelogram. Is it a rectangle?
Solution: Midpoint of AC = ((−2 + 4)/2, (−1 + 3)/2) = (1, 1). Midpoint of BD = ((1 + 1)/2, (0 + 2)/2) = (1, 1).
The diagonals bisect each other, so ABCD is a parallelogram (Test 3).
AC = √(62 + 42) = √52 and BD = √(02 + 22) = 2. The diagonals are unequal, so ABCD is not a rectangle.
Example 11: Midpoint of the hypotenuse
△ABC is right-angled at C. A line through the midpoint M of hypotenuse AB, parallel to BC, meets AC at D. Prove that (i) D is the midpoint of AC, (ii) MD ⊥ AC, (iii) CM = MA = AB/2.
Solution: (i) M is the midpoint of AB and MD ∥ BC, so by the converse of the midpoint theorem, D is the midpoint of AC.
(ii) MD ∥ BC with transversal AC, so ∠ADM = ∠ACB = 90° (corresponding angles). So MD ⊥ AC.
(iii) In △ADM and △CDM: AD = CD (from (i)), ∠ADM = ∠CDM = 90°, and DM is common. So △ADM ≅ △CDM (SAS) and CM = AM. Since M is the midpoint of AB, CM = MA = AB/2.
Example 12: Trisecting a diagonal
In parallelogram ABCD, E and F are the midpoints of AB and CD. Prove that AF and EC trisect the diagonal BD.
Solution: Let AF meet BD at P and EC meet BD at Q.
AE = AB/2 and CF = CD/2, and AB = CD, so AE = CF. Also AE ∥ CF (parts of AB ∥ DC). One pair of opposite sides of AECF is equal and parallel, so AECF is a parallelogram and AF ∥ EC.
In △DQC, F is the midpoint of DC and FP ∥ CQ. By the converse of the midpoint theorem, P is the midpoint of DQ, so DP = PQ.
In △BPA, E is the midpoint of AB and EQ ∥ AP. So Q is the midpoint of BP, and BQ = PQ.
So DP = PQ = QB, and BD is divided into three equal parts.
Competency-Based Questions (with answers)
1. Case-based: The folding gate
A shop has a folding metal gate made of strips joined by pins. Each unit of the gate is a 4-gon ABCD in which AB = CD = 40 cm and AD = BC = 25 cm, and the pins let the angles change as the gate opens and closes.
(a) What shape is each unit, whatever the angles? A parallelogram, because both pairs of opposite sides are equal (Test 1).
(b) When the gate is half open, ∠A = 55°. Find the other angles. ∠B = 180° − 55° = 125°, ∠C = 55° and ∠D = 125°.
(c) At one position the two diagonals of a unit are equal. What shape is the unit then? A parallelogram with equal diagonals is a rectangle, so all four angles are 90°.
2. Assertion-Reason
Assertion (A): A 4-gon whose diagonals are equal is a rectangle. Reason (R): The diagonals of a rectangle are equal.
Options: (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). R is a true theorem. A is its converse, and it is false: an isosceles trapezium has equal diagonals and is not a rectangle. Equal diagonals give a rectangle only when the 4-gon is already known to be a parallelogram.
3. Assertion-Reason
Assertion (A): If the diagonals of a 4-gon bisect each other at right angles, the 4-gon is a rhombus. Reason (R): A parallelogram whose diagonals are perpendicular is a rhombus.
Options: (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). Diagonals that bisect each other make the 4-gon a parallelogram (Test 3). R then applies, because those diagonals are perpendicular, so the parallelogram is a rhombus.
4. Assertion-Reason
Assertion (A): In △ABC, D and E are the midpoints of AB and AC. If DE = 6 cm, then BC = 12 cm. Reason (R): The medians of a triangle divide each other in the ratio 2 : 1.
Options: (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: (b). A is true by the midpoint theorem, BC = 2DE = 12 cm. R is a true fact about the centroid, but it plays no part in finding BC.
5. Error analysis: Aman’s proof
Aman writes: “In 4-gon ABCD, AB ∥ DC and AD = BC. So ABCD is a parallelogram.”
Find the error. One pair of sides parallel and the other pair equal is not a valid test. An isosceles trapezium, for example with parallel sides AB = 10 cm and DC = 4 cm and equal legs AD = BC = 5 cm, satisfies both conditions and is not a parallelogram. A correct test needs the same pair of sides to be both equal and parallel (Test 4).
6. Coordinate case: The park plan
On a town map, three corners of a parallelogram-shaped park are at P(2, 1), Q(7, 2) and R(9, 6), taken in order round the park.
(a) Find the fourth corner S. S is opposite Q, so S = (2 + 9 − 7, 1 + 6 − 2) = (4, 5).
(b) A fountain is placed where the paths along the two diagonals cross. Find its position. At the common midpoint of PR and QS: ((2 + 9)/2, (1 + 6)/2) = (5.5, 3.5). Check with QS: ((7 + 4)/2, (2 + 5)/2) = (5.5, 3.5).
7. Paper-folding activity: Midpoints of a 4-gon
Riya cuts out an irregular 4-gon, marks the midpoints of its four sides by folding each side end to end, and joins them in order. She measures the new 4-gon and finds that its opposite sides are equal.
(a) What shape will she always get? A parallelogram, for any 4-gon, by the midpoint 4-gon theorem.
(b) She repeats this on a rhombus. What does she get, and why? A rectangle. The sides of the midpoint 4-gon are parallel to the diagonals of the rhombus, and those diagonals are perpendicular, so adjacent sides of the new 4-gon are perpendicular.
(c) The diagonals of her original 4-gon are 14 cm and 11 cm. Find the perimeter of the midpoint 4-gon. Perimeter = sum of the diagonals = 14 + 11 = 25 cm.
8. Reasoning: Forming converses
Write the converse of “If a 4-gon is a square, then its diagonals are equal and perpendicular”, and decide whether it is true.
Answer: Converse: “If the diagonals of a 4-gon are equal and perpendicular, then it is a square.” It is false. Take a kite ABCD whose diagonals AC and BD are each 10 cm long and perpendicular, meeting at O with BO = OD = 5 cm, AO = 3 cm and OC = 7 cm. The diagonals do not bisect each other, so it is not even a parallelogram. The true statement needs the diagonals also to bisect each other.
Important Questions for Exams
1-Mark Questions
- Three angles of a 4-gon are 90°, 90° and 100°. Find the fourth angle. 360° − 280° = 80°.
- One angle of a parallelogram is 65°. Find the angle adjacent to it. 180° − 65° = 115°.
- In △ABC, D and E are the midpoints of AB and AC and BC = 9 cm. Find DE. DE = 9/2 = 4.5 cm.
- How many lines of symmetry does a rhombus that is not a square have? Two, along its diagonals.
- Find the midpoint of the segment joining (−3, 4) and (5, −2). ((−3 + 5)/2, (4 − 2)/2) = (1, 1).
2-Mark Questions
- Can a 4-gon have angles 110°, 80°, 70° and 95°? Give a reason. No. The sum is 355°, and the angles of a 4-gon must add to 360°.
- Show that the diagonals of a rhombus are perpendicular to each other. Let the diagonals of rhombus ABCD meet at O. In △AOB and △AOD, AB = AD, OB = OD (diagonals bisect each other) and AO is common, so the triangles are congruent (SSS). Then ∠AOB = ∠AOD, and they form a linear pair, so each is 90°.
- The median AD of △ABC is 9 cm. Find the distance of the centroid from A and from D. AG = (2/3) × 9 = 6 cm and GD = 3 cm.
- Three vertices of parallelogram ABCD are A(3, −1), B(5, 2) and C(2, 4). Find D. D = (3 + 2 − 5, −1 + 4 − 2) = (0, 1).
3-Mark Questions
- Prove that a diagonal of a parallelogram divides it into two congruent triangles. In parallelogram ABCD with diagonal AC: ∠BAC = ∠DCA (alternate angles, AB ∥ DC), ∠BCA = ∠DAC (alternate angles, AD ∥ BC), and AC is common. So △ABC ≅ △CDA by ASA.
- Show that if the diagonals of a 4-gon bisect each other, it is a parallelogram. Let AC and BD meet at O with OA = OC and OB = OD. △AOB ≅ △COD (SAS, with vertically opposite angles at O), so ∠OAB = ∠OCD, alternate angles, giving AB ∥ DC. Similarly △AOD ≅ △COB gives AD ∥ BC. So ABCD is a parallelogram.
- ABCD is a rectangle and P, Q, R, S are the midpoints of AB, BC, CD, DA. Show that PQRS is a rhombus. By the midpoint theorem in △ABC and △ADC, PQ = SR = AC/2 and PQ ∥ SR, so PQRS is a parallelogram. In △ABD, PS = BD/2. The diagonals of a rectangle are equal, so AC = BD and PQ = PS. A parallelogram with two adjacent sides equal is a rhombus.
5-Mark Questions
- State and prove the midpoint theorem. Use it to find the perimeter of the triangle formed by joining the midpoints of the sides of a triangle with sides 7 cm, 9 cm and 12 cm. Statement: the segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it. Proof: with E, F the midpoints of AB, AC, extend EF to D so that FD = EF; △AEF ≅ △CDF (SAS), so CD = AE = EB and CD ∥ EB (alternate angles ∠EAF = ∠DCF); EBCD is a parallelogram (Test 4), so EF ∥ BC and EF = ED/2 = BC/2. The midpoint triangle has sides 3.5, 4.5 and 6 cm, perimeter 14 cm.
- Prove that the 4-gon formed by joining the midpoints of the sides of a 4-gon, in order, is a parallelogram. When is it a rectangle? With P, Q, R, S the midpoints of AB, BC, CD, DA: in △ABC, PQ ∥ AC and PQ = AC/2; in △ADC, SR ∥ AC and SR = AC/2. So PQ = SR and PQ ∥ SR, and PQRS is a parallelogram by Test 4. PQ ∥ AC and PS ∥ BD, so ∠SPQ = 90° exactly when AC ⊥ BD. The midpoint 4-gon is a rectangle when the diagonals of ABCD are perpendicular, as in a rhombus or a kite.
Common Mistakes and Examiner Tips
- Using a converse without proof. A 4-gon with equal diagonals need not be a rectangle. Use only the proved tests.
- Writing “opposite sides parallel” as a test when only one pair is known. One parallel pair gives a trapezium. You need both pairs parallel, or one pair both equal and parallel.
- Mixing up which pair is equal and parallel. Test 4 needs the SAME pair of sides to be equal and parallel. AB ∥ DC with AD = BC is an isosceles trapezium pattern.
- Assuming a parallelogram has equal or perpendicular diagonals. A general parallelogram has neither. Equal diagonals belong to rectangles and perpendicular diagonals to rhombuses.
- Taking a diagonal of a rectangle as a line of symmetry. The lines of symmetry of a non-square rectangle join the midpoints of opposite sides.
- Getting the 2 : 1 ratio the wrong way round. The longer part of a median is next to the vertex: AG = (2/3)AD, GD = (1/3)AD. Write the ratio as vertex part : side part.
- Doubling instead of halving in the midpoint theorem. The midpoint segment is half the third side. If you know the midpoint segment, the third side is twice it.
- Wrong vertex order in the fourth-vertex formula. In ABCD, D is opposite B, so D = A + C − B. Using A + B − C gives the parallelogram ACBD, in which the new point is opposite C, a different figure.
- Skipping reasons in proofs. Give every step a reason in brackets: alternate angles (naming the parallel lines), vertically opposite angles, common side, CPCT. Write congruences in matching vertex order, such as △ABC ≅ △CDA.
- Using 180° as the angle sum of a 4-gon. The sum is 360°. In a parallelogram, adjacent angles add to 180°.
Quick Revision Points
- A 4-gon: four points, no three collinear, joined in order by sides that meet only at their ends
- Angle sum of any 4-gon = 360°; at most three of its angles can be obtuse
- A diagonal splits a parallelogram into two congruent triangles (ASA)
- Parallelogram: opposite sides equal, opposite angles equal, adjacent angles supplementary
- Tests: both pairs of opposite sides equal, or both pairs of opposite angles equal, or diagonals bisect each other, or one pair of sides equal and parallel
- Rectangle: all angles 90°, diagonals equal; a parallelogram with equal diagonals is a rectangle
- Rhombus: all sides equal, diagonals perpendicular bisectors of each other and angle bisectors
- Square: rectangle and rhombus together; 4 lines of symmetry, rotation by 90°
- A 4-gon has central symmetry exactly when it is a parallelogram; the centre is where the diagonals meet
- Any 4-gon tiles the plane: the four angles meet at each corner and add to 360°
- Midpoint theorem: EF ∥ BC and EF = BC/2
- Midpoints of the sides of any 4-gon form a parallelogram, with perimeter AC + BD
- Midpoint 4-gon of a rhombus is a rectangle; of a rectangle is a rhombus; of a square is a square
- Medians are concurrent at the centroid G; AG : GD = 2 : 1
- Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
- Fourth vertex of ABCD: D = (xA + xC − xB, yA + yC − yB)
Weightage and Exam Pattern
Class 9 has no board exam. Quadrilaterals is assessed in school exams: periodic tests, the half-yearly exam and the annual exam. In the CBSE Class IX Mathematics course structure for 2026-27, the chapter appears as “4-gons (Quadrilaterals)” in Unit IV, Geometry, which carries 25 marks out of the 80-mark theory paper and is allotted 69 teaching periods. The course structure does not split the 25 marks between the five Geometry chapters. The other 20 marks are internal assessment.
| 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 CBSE question paper design weights questions by typology: about 54% remembering and understanding, 24% applying, and 22% analysing, evaluating and creating. Here that means properties of special 4-gons, calculations with parallelograms, midpoints and medians, and proofs such as the parallelogram tests and the midpoint theorem.
The course structure also expects drawing, measurement and paper manipulation activities to discover patterns involving triangles and 4-gons, such as the paper-folding task in Competency Question 7. Internal assessment includes a 5-mark lab practical based on activities from the prescribed books. The midpoint theorem and the parallelogram tests return in the Class 10 chapters on triangles and coordinate geometry.
Class 9 Maths – swipe through all 10 cards to understand the whole chapter.
What a 4-gon Is
Four points, no three on a line, joined in order by sides that meet only at their ends.
A diagonal inside splits it into two triangles: 2 × 180°.
- Angles in ratio 3 : 5 : 9 : 13 give 36°, 60°, 108°, 156°.
- At most three angles can be obtuse; all four cannot be acute.
- A 4-gon frame is not rigid; a diagonal brace makes two rigid triangles.
Parallelogram Properties
A diagonal splits a parallelogram into two congruent triangles by ASA.
Adjacent angles are co-interior, so ∠A + ∠B = 180°.
- Diagonals bisect each other: OA = OC, OB = OD.
- One angle 70° means the angles are 70°, 110°, 70°, 110°.
- Diagonals are in general neither equal nor perpendicular.
Four Tests for a Parallelogram
Any one of these conditions proves a 4-gon is a parallelogram.
One pair parallel plus the other pair equal is NOT a test.
- Both pairs of opposite sides equal (SSS).
- Both pairs of opposite angles equal (angle sum).
- One pair of opposite sides equal and parallel (SAS).
Rectangle, Rhombus, Square
Special parallelograms add a condition on the angles, the sides or the diagonals.
A square is both a rectangle and a rhombus.
- Rectangle: diagonals equal; parallelogram with equal diagonals is a rectangle.
- Rhombus: diagonals are perpendicular bisectors and bisect the vertex angles.
- Diagonals 24 cm and 10 cm give a rhombus side of 13 cm.
Central Symmetry and Tiling
A half-turn about the meeting point of the diagonals maps a parallelogram onto itself.
Any 4-gon tiles a plane: its four angles meet at each corner and add to 360°.
- General parallelogram: 0 lines of symmetry, half-turn symmetry.
- Rectangle and rhombus: 2 lines each; square: 4 lines and a 90° turn.
- Kite and isosceles trapezium: 1 line, no half-turn symmetry.
Midpoint Theorem
The segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it.
Proof: extend EF to D with FD = EF, then EBCD is a parallelogram.
- Converse: a line through a midpoint, parallel to a side, bisects the third side.
- Midpoint triangle of a triangle with sides 8, 10, 12 has perimeter 15.
- The three midpoint segments cut a triangle into four congruent triangles.
The Midpoint 4-gon
Joining the midpoints of the sides of any 4-gon in order always gives a parallelogram.
PQ and SR are both parallel to AC and equal to half of it.
- Diagonals perpendicular (rhombus, kite) give a rectangle.
- Diagonals equal (rectangle) give a rhombus.
- A square gives a square.
Medians and the Centroid
The three medians of a triangle meet at the centroid G.
AG = (2/3)AD and GD = (1/3)AD.
- Proof uses the midpoint theorem twice to build a parallelogram.
- Median 15 cm gives AG = 10 cm, GD = 5 cm.
- The longer part of a median is next to the vertex.
Midpoints and Fourth Vertex
The diagonals of a parallelogram share a midpoint, which turns geometry into arithmetic.
In ABCD, D = (xA + xC − xB, yA + yC − yB).
- A(1, 2), B(4, 3), C(6, 6) give D = (3, 5).
- Same midpoint for both diagonals proves a parallelogram.
- Add equal diagonals (distance formula) to prove a rectangle.
Converses That Fail
A true theorem can have a false converse; one counter example is enough to show it.
Always check a converse against an isosceles trapezium or a kite.
- Equal diagonals alone do not give a rectangle (isosceles trapezium).
- Equal and perpendicular diagonals alone do not give a square (kite).
- AB ∥ DC with AD = BC need not be a parallelogram.
📝 Practice Quadrilaterals - 10 questions
Concept-check questions · with answers & explanations
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Why A: Step 1 (angle sum of a 4-gon): angle A + angle B = 360° – 100° – 60° = 200°. Step 2 (angle bisectors halve the angles): angle PAB + angle PBA = 200° / 2 = 100°. Step 3 (angle sum of triangle APB): angle APB = 180° – 100° = 80°. This equals (angle C + angle D) / 2.
Why not B: 100° is the sum of the two half angles PAB and PBA, not the third angle of the triangle.
Why not C: 160° is angle C + angle D; the bisector angle is half of this.
Why not D: 40° halves the 80° again; the angle is (C + D) / 2, not (C + D) / 4.
Remember: The angle between the bisectors of A and B equals (C + D) / 2.
Why D: Step 1 (angle bisector): angle DAE = angle EAB. Step 2 (alternate angles, AB ∥ DC with transversal AE): angle EAB = angle AED. So angle DAE = angle AED. Step 3 (sides opposite equal angles are equal): in triangle ADE, DE = AD = 6 cm. Step 4 (opposite sides of a parallelogram): DC = AB = 10 cm, so EC = 10 – 6 = 4 cm.
Why not A: 6 cm is DE, the part next to D, not EC.
Why not B: 10 cm is the whole side DC.
Why not C: 16 cm adds AB and AD; the parts of DC must add up to DC = 10 cm.
Remember: Angle bisector plus parallel sides gives an isosceles triangle.
Why D: The Assertion is false: an isosceles trapezium ABCD with AB ∥ DC and AD = BC has one pair of opposite sides equal and the other pair parallel, but AB and DC have different lengths, so it is not a parallelogram. The Reason is a true theorem, proved by SSS congruence on a diagonal. So A is false and R is true.
Why not A: A is false because the isosceles trapezium is a counterexample.
Why not B: Both cannot be true, since the isosceles trapezium disproves A.
Why not C: This reverses the truth values: R is the true theorem and A is the false statement.
Remember: Mixed conditions (one pair equal, other pair parallel) fail: think isosceles trapezium.
Why D: Each diagonal of a rhombus is the perpendicular bisector of the other and bisects the vertex angles, so folding along either diagonal matches the halves: 2 lines of symmetry. Like every parallelogram, it also has half-turn symmetry about the centre. A quarter-turn does not work unless all angles are 90°.
Why not A: 4 lines and a 90° turn belong only to a square.
Why not B: Lines through midpoints of opposite sides are the fold lines of a rectangle, not a rhombus.
Why not C: That describes a general parallelogram; equal sides give a rhombus two fold lines.
Remember: Rhombus folds on its diagonals, rectangle folds through side midpoints.
Why C: By the midpoint theorem, PQ ∥ AC and QR ∥ BD, and PQRS is a parallelogram. The diagonals of a rhombus are perpendicular, so PQ ⊥ QR. A parallelogram with a right angle is a rectangle. (It is a square only when the rhombus is itself a square.)
Why not A: A rhombus midpoint figure arises when the diagonals are EQUAL, as in a rectangle; a rhombus has perpendicular diagonals instead.
Why not B: A square needs the diagonals of ABCD to be equal as well as perpendicular, which holds only if ABCD is a square.
Why not D: PQRS has both pairs of opposite sides parallel, so it is a parallelogram, not a kite.
Remember: Perpendicular diagonals give a rectangle, equal diagonals give a rhombus.
Why A: Step 1 (midpoint formula): the midpoint of BC is D = ((-4 + 4)/2, (0 + 0)/2) = (0, 0), so the median AD runs from (0, 6) to (0, 0) and has length 6. Step 2 (centroid divides the median 2 : 1 from the vertex): AG = (2/3) × 6 = 4, so G is 4 units below A: G = (0, 6 – 4) = (0, 2).
Why not B: (0, 4) is two thirds of the way from D, not from A; the ratio is reversed.
Why not C: (0, 3) is the midpoint of the median AD.
Why not D: (0, 0) is D, the midpoint of BC, the foot of the median.
Remember: Find the median’s foot, then go two thirds from the vertex.
Why D: When copies are placed by rotating through 180° about the midpoints of sides, one copy of each of the angles A, B, C and D meets at every corner point. Their sum is the angle sum of the 4-gon, 360°, which is exactly one full turn, so they fill the space around the point with no gap and no overlap.
Why not A: Only parallelograms have two pairs of parallel sides; an irregular 4-gon usually has none, yet it still tiles.
Why not B: Only rectangles have all angles 90°; the tiling works for any angles because only their SUM matters.
Why not C: Diagonals are equal only in special cases such as rectangles and isosceles trapeziums, and tiling does not depend on diagonals.
Remember: Four different angles meet at each point and fill 360°, one full turn.
Why B: Step 1 (diagonals of a parallelogram bisect each other): OA = OC, so 3x – 1 = x + 5, giving 2x = 6 and x = 3. Step 2 (substitute): OA = 3(3) – 1 = 8 cm. Step 3 (AC is twice OA): AC = 2 × 8 = 16 cm.
Why not A: 8 cm is OA, only half of the diagonal.
Why not C: 3 is the value of x, not a length.
Why not D: 10 cm follows from 2x = 4, a sign slip, giving x = 2 and OA = 5.
Remember: O is the midpoint of both diagonals, so AC = 2 × OA.
Why D: A rectangle has equal diagonals, but the converse fails: an isosceles trapezium also has equal diagonals and is not a rectangle (not even a parallelogram). One counterexample is enough to show a converse is false. The other three statements are proved characterisations of a parallelogram.
Why not A: This is true, proved by SAS on the triangles at O.
Why not B: This is true: the angles then give angle A + angle B = 180°, so the sides are parallel.
Why not C: This is true, proved by SSS on a diagonal.
Remember: Not every converse is true: the isosceles trapezium breaks the equal-diagonals converse.
Why A: Step 1 (diagonals of a rectangle are equal and bisect each other): OA = OB, so triangle OAB is isosceles and angle OBA = 32°. Step 2 (angle sum of triangle OAB): angle AOB = 180° – 32° – 32° = 116°. Step 3 (linear pair along BD): angle AOD = 180° – 116° = 64°.
Why not B: 116° is angle AOB, not angle AOD.
Why not C: 58° is angle OAD = 90° – 32°, an angle at A, not at O.
Why not D: 32° is angle OAB itself; the angle at O is found through the isosceles triangle.
Remember: In a rectangle OA = OB = OC = OD, so look for isosceles triangles.
Chapter Navigation
Previous: Sequences and Progressions Class 9 Notes
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- Coordinate Geometry Class 9 Notes
- Introduction to Polynomials Class 9 Notes
- Number System Class 9 Notes
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- Circles Class 9 Notes
- Area and Perimeter Class 9 Notes
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Frequently Asked Questions
The four angles of any 4-gon add to 360°. Draw a diagonal that lies inside the 4-gon. It splits the figure into two triangles, and the six angles of those triangles make up exactly the four angles of the 4-gon. Each triangle has angle sum 180°, so the total is 2 × 180° = 360°.
Use any one of these tests: both pairs of opposite sides are equal; both pairs of opposite angles are equal; the diagonals bisect each other; or one pair of opposite sides is both equal and parallel. The side and diagonal tests work by finding two congruent triangles and then equal alternate angles, which force the sides to be parallel. The angle test uses the 360° angle sum and co-interior angles.
No. An isosceles trapezium has equal diagonals and is not a rectangle. The correct statement is that a parallelogram with equal diagonals is a rectangle. This is a standard example of a converse that is false until an extra condition is added, here that the 4-gon is already a parallelogram.
The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it. So if E and F are the midpoints of AB and AC in triangle ABC, then EF ∥ BC and EF = BC/2. The proof extends EF to D with FD = EF and shows EBCD is a parallelogram.
A line drawn through the midpoint of one side of a triangle, parallel to a second side, bisects the third side. It is proved by taking the true midpoint of the third side, applying the midpoint theorem, and using the fact that only one line through a point can be parallel to a given line.
The three medians of a triangle meet at one point, the centroid G, which divides each median in the ratio 2 : 1 from the vertex. If median AD is 12 cm, then AG = 8 cm and GD = 4 cm. The longer part is always the one next to the vertex.
The diagonals of parallelogram ABCD bisect each other, so AC and BD have the same midpoint. This gives D = (xA + xC − xB, yA + yC − yB). For A(1, 2), B(4, 3), C(6, 6), D = (1 + 6 − 4, 2 + 6 − 3) = (3, 5). Keep the order of letters: D is opposite B.
