Quadrilaterals

1. Quadrilateral basics and the angle-sum property

A quadrilateral is a closed plane figure formed by four line segments called sides, meeting at four points called vertices. The line segments joining opposite vertices are called diagonals.

Naming: Quadrilateral ABCD has vertices A, B, C, D in order. Sides: AB, BC, CD, DA; diagonals: AC and BD. Opposite sides: (AB, CD) and (BC, DA). Adjacent angles share a side — e.g. A and B are adjacent.

Convex vs. Concave: A quadrilateral is convex if both diagonals lie entirely inside it. It is concave if one diagonal lies outside. This chapter deals only with convex quadrilaterals.

Theorem 8.1 — Angle Sum Property

Statement: The sum of the angles of a quadrilateral is 360°.

Proof: Let ABCD be a quadrilateral. Draw diagonal AC, dividing it into triangles ABC and ACD.

• In triangle ABC: $\angle BAC + \angle ABC + \angle BCA = 180°$
• In triangle ACD: $\angle DAC + \angle ACD + \angle CDA = 180°$

Adding both equations and noting that $\angle BAC + \angle DAC = \angle DAB$ and $\angle BCA + \angle ACD = \angle BCD$:

Two triangles, each contributing 180°, give $2 \times 180° = 360°.$ $\blacksquare$

2. Types of quadrilaterals — hierarchy and definitions

Quadrilaterals are classified by how many pairs of parallel or equal sides they have.

Hierarchy — read as "is a special case of":

Every square is both a rectangle and a rhombus. Every rectangle and rhombus is a parallelogram. Every parallelogram is a trapezium (with two pairs of parallel sides). A kite is generally not a parallelogram.

Isosceles trapezium: A trapezium whose non-parallel sides (legs) are equal. Its base angles are equal, and its diagonals are equal.

3. Properties of a parallelogram — three theorems with proofs

In all proofs, ABCD is a parallelogram: AB ∥ CD and AD ∥ BC.

Theorem 8.2 — Opposite sides are equal

Statement: If ABCD is a parallelogram, then AB = CD and AD = BC.

Proof: Draw diagonal AC.

• In $\triangle ABC$ and $\triangle CDA$:
• $\angle BAC = \angle DCA$  (alternate interior angles, AB ∥ CD, transversal AC)
• $\angle BCA = \angle DAC$  (alternate interior angles, AD ∥ BC, transversal AC)
• $AC = CA$  (common side)
• By ASA: $\triangle ABC \cong \triangle CDA$

Therefore $AB = CD$ and $BC = DA.$ $\blacksquare$

Converse (Theorem 8.5): If opposite sides of a quadrilateral are equal, it is a parallelogram.

Theorem 8.3 — Opposite angles are equal

Statement: If ABCD is a parallelogram, then $\angle A = \angle C$ and $\angle B = \angle D.$

Proof: Since AB ∥ CD, co-interior angles with transversal AD give $\angle A + \angle D = 180°.$ Since AD ∥ BC, co-interior angles with transversal AB give $\angle A + \angle B = 180°.$ Therefore $\angle B = \angle D.$ Similarly, using transversals BC and CD gives $\angle A = \angle C.$ $\blacksquare$

Alternatively, from Theorem 8.2's congruence: $\triangle ABC \cong \triangle CDA \Rightarrow \angle B = \angle D.$ Then angle sum gives $\angle A = \angle C.$ $\blacksquare$

Note: Adjacent angles of a parallelogram are supplementary (sum = 180°), since they are co-interior angles between parallel lines.

Converse (Theorem 8.6): If opposite angles of a quadrilateral are equal, it is a parallelogram.

Theorem 8.4 — Diagonals bisect each other

Statement: If ABCD is a parallelogram with diagonals AC and BD intersecting at O, then OA = OC and OB = OD.

Proof: In $\triangle AOB$ and $\triangle COD$:

• $\angle OAB = \angle OCD$  (alternate interior angles, AB ∥ CD)
• $AB = CD$  (opposite sides, Theorem 8.2)
• $\angle OBA = \angle ODC$  (alternate interior angles, AB ∥ CD)
• By ASA: $\triangle AOB \cong \triangle COD$

Therefore $OA = OC$ and $OB = OD.$ $\blacksquare$

Converse (Theorem 8.7): If the diagonals of a quadrilateral bisect each other, it is a parallelogram.

4. Conditions for a quadrilateral to be a parallelogram

Any ONE of the following conditions is sufficient to guarantee the quadrilateral is a parallelogram:

Theorem 8.8 — One pair of sides equal and parallel

Statement: If one pair of opposite sides of a quadrilateral is both equal and parallel, then it is a parallelogram.

Proof sketch: Let ABCD have AB ∥ CD and AB = CD. Draw diagonal AC. In $\triangle ABC$ and $\triangle CDA$: AB = CD (given), $\angle BAC = \angle DCA$ (alternate angles, AB ∥ CD), AC = CA (common). By SAS: $\triangle ABC \cong \triangle CDA,$ so $AD = CB$ and $\angle DAC = \angle BCA.$ These are alternate interior angles, so AD ∥ BC. Both pairs of opposite sides are now parallel. $\blacksquare$

5. Rectangle, rhombus and square as special parallelograms

All three are parallelograms and inherit every parallelogram property. Each adds one or two extra conditions.

Rectangle

Definition: A parallelogram in which each angle is 90°.

Extra property — diagonals are equal:

Theorem: The diagonals of a rectangle are equal.

Proof: In rectangle ABCD, $\angle A = \angle B = 90°.$ In $\triangle ABD$ and $\triangle BAC$: AB = BA (common), $\angle DAB = \angle CBA = 90°,$ AD = BC (opp. sides of parallelogram). By SAS: $\triangle ABD \cong \triangle BAC,$ so BD = AC. $\blacksquare$

Converse: If the diagonals of a parallelogram are equal, it is a rectangle.

Key facts: Rectangle has 4 right angles; diagonals equal and bisect each other (but NOT perpendicular unless it is also a square).

Rhombus

Definition: A parallelogram in which all four sides are equal.

Extra property — diagonals are perpendicular bisectors: (proved above in Example 3)

Converse: If the diagonals of a parallelogram bisect each other at right angles, it is a rhombus.

Diagonal angle bisector property: Each diagonal of a rhombus bisects the vertex angles through which it passes.

Key facts: Rhombus has 4 equal sides; diagonals bisect each other at 90° and bisect the vertex angles; opposite angles equal, but NOT generally 90°.

Square

Definition: A parallelogram that is simultaneously a rectangle and a rhombus (4 equal sides AND all angles 90°).

Properties: Diagonals are equal, perpendicular, bisect each other, and bisect the vertex angles (each 90° vertex angle is bisected into 45°).

6. Mid-Point Theorem — statement and full proof

This is the most important theorem in the chapter for higher-order problems and board proofs.

Theorem 8.9 — The Mid-Point Theorem

Statement: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.

Given: Triangle ABC with E and F the mid-points of AB and AC respectively (so AE = EB and AF = FC).

To prove: EF ∥ BC and $EF = \tfrac{1}{2}BC.$

Construction: Extend EF to G such that EF = FG. Join CG.

Proof:

1. In $\triangle AEF$ and $\triangle CGF$:
   • $AF = CF$ (F is mid-point of AC — given)
   • $\angle AFE = \angle CFG$ (vertically opposite angles)
   • $EF = GF$ (by construction)
   • By SAS: $\triangle AEF \cong \triangle CGF$
2. From CPCT: $AE = CG$ and $\angle AEF = \angle CGF.$ Since $\angle AEF$ and $\angle CGF$ are alternate interior angles for lines AB and CG with transversal EG, we get $AE$ ∥ $CG,$ i.e. $EB$ ∥ $CG.$
3. Also $EB = AE = CG$ (since AE = EB because E is mid-point, and AE = CG from CPCT). So in quadrilateral EBCG: $EB$ ∥ $CG$ and $EB = CG.$ By Theorem 8.8, EBCG is a parallelogram.
4. Since EBCG is a parallelogram, $EG$ ∥ $BC,$ i.e. $EF$ ∥ $BC.$ And $BC = EG = 2 \times EF,$ so $EF = \tfrac{1}{2}BC.$ $\blacksquare$

7. Converse of the Mid-Point Theorem

Theorem 8.10 — Converse of the Mid-Point Theorem

Statement: The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

Given: Triangle ABC. E is the mid-point of AB. EF ∥ BC, where F lies on AC.

To prove: F is the mid-point of AC (i.e. AF = FC).

Proof: Construct G on extension of EF such that EF = FG. Join CG. By the congruence argument identical to Theorem 8.9, $\triangle AEF \cong \triangle CGF$ (SAS: AF = CF — to prove), giving EBCG as a parallelogram. Since EBCG is a parallelogram, $EG$ ∥ $BC.$ But we are given $EF$ ∥ $BC.$ So G lies on line EF. The CPCT from the congruence then gives $AF = CF,$ i.e. F is the mid-point of AC. $\blacksquare$

Classic exam application

8. Exercise solutions — key questions

Ex 8.1 — Q1: Three angles 75°, 90°, 75°. Fourth angle $= 360° - (75°+90°+75°) = 120°.$

Ex 8.1 — Q2: If the diagonals of a parallelogram are equal, show it is a rectangle. In $\triangle ABD$ and $\triangle BAC$: $AD = BC$ (opp. sides), $AB = BA$ (common), $BD = AC$ (given). By SSS: $\triangle ABD \cong \triangle BAC,$ so $\angle DAB = \angle CBA.$ But they are co-interior angles: $\angle DAB + \angle CBA = 180°.$ So each $= 90°.$ A parallelogram with a right angle is a rectangle. $\blacksquare$

Ex 8.1 — Q3: If diagonals of a quadrilateral bisect each other at 90°, show it is a rhombus. $OA = OC,$ $OB = OD,$ $\angle AOB = 90°.$ In each of the four triangles formed (AOB, BOC, COD, DOA), SAS congruence (equal half-diagonals and 90°) gives $AB = BC = CD = DA.$ Diagonals bisect each other means it is a parallelogram; equal sides means it is a rhombus. $\blacksquare$

Ex 8.1 — Q6: Diagonal AC of parallelogram ABCD bisects $\angle A.$ Show AC bisects $\angle C$ and ABCD is a rhombus. (Full solution in Example 6 above.) $\blacksquare$

Ex 8.2 — Q1: ABCD is a quadrilateral; E, F, G, H are mid-points of AB, BC, CD, DA. Show EFGH is a parallelogram. Draw diagonal AC. By mid-point theorem in $\triangle ABC$: $EF$ ∥ $AC$ and $EF = \tfrac{1}{2}AC.$ By mid-point theorem in $\triangle ACD$: $HG$ ∥ $AC$ and $HG = \tfrac{1}{2}AC.$ So $EF$ ∥ $HG$ and $EF = HG.$ EFGH is a parallelogram. $\blacksquare$

Ex 8.2 — Q2: Rhombus ABCD with AC = 24 cm, BD = 10 cm. Find the perimeter. Diagonals bisect at right angles: $OA = 12,$ $OB = 5.$ $AB = \sqrt{12^2+5^2} = \sqrt{169} = 13$ cm. Perimeter $= 4 \times 13 = 52$ cm.

Ex 8.2 — Q4: P, Q, R, S are mid-points of sides of rectangle ABCD. Show PQRS is a rhombus. Since rectangle diagonals AC = BD and mid-point theorem gives $PQ = QR = RS = SP = \tfrac{1}{2}AC = \tfrac{1}{2}BD.$ All sides of PQRS are equal and PQRS is a parallelogram (proved like Q1), so PQRS is a rhombus. $\blacksquare$

Ex 8.2 — Q7: ABC and BCD are triangles on a common base BC. If E is mid-point of AC, then prove $DE$ ∥ $AB.$ In $\triangle ACD$ (or using the converse of mid-point theorem): if E is mid-point of AC and we need $DE$ ∥ $AB,$ draw median from D. Using the converse, if the line through E is parallel to BC in a related triangle, it bisects the third side. The exact figure context determines the proof path; the converse of mid-point theorem is the key tool. $\blacksquare$

9. Common mistakes to avoid

• Angle sum: Writing 180° instead of 360° — 360° is the quadrilateral sum; 180° is the triangle sum.
• Diagonals of a rectangle: Calling them perpendicular — they are perpendicular only in a square.
• Diagonals of a rhombus: Calling them equal — they are equal only in a square.
• Mid-point theorem direction: The line through mid-points of AB and AC is parallel to BC (the third side) and half of BC.
• One pair of sides equal: That is NOT enough to guarantee a parallelogram. You need BOTH equal AND parallel (Th. 8.8), or both pairs equal (Th. 8.5).
• Proof writing: State the congruence rule (SAS, ASA, SSS, AAS) and the reason for each angle equality (alternate interior, co-interior, vertically opposite). Missing reasons lose marks.
• CPCT: Always write "by CPCT" when deriving side or angle equalities after proving congruence — examiners look for this phrase.