Quadrilaterals

A quadrilateral is a closed plane figure bounded by four line segments (sides), with four vertices and four interior angles. Squares, rectangles, kites and general four-sided shapes are all quadrilaterals.

To find the sum of its four angles, draw one diagonal of the quadrilateral ABCD, say AC. This diagonal divides ABCD into two triangles, ΔABC and ΔACD. Every angle of the quadrilateral is used exactly once between these two triangles (the diagonal splits ∠A and ∠C into two parts each, which add back up to the full angle).

Since the angle sum of a triangle is 180°, adding the two triangles gives:

∠A + ∠B + ∠C + ∠D = 180° + 180° = 360°

This is true for every quadrilateral, whether it is convex, a parallelogram, a kite, or an irregular four-sided shape — as long as it can be split into two triangles by a diagonal.

Worked Example: Three angles of a quadrilateral are 80°, 100° and 75°. Find the fourth angle.

Let the fourth angle be x. By the angle sum property:

80° + 100° + 75° + x = 360°

255° + x = 360°

x = 360° − 255° = 105°

Not every quadrilateral looks alike. Based on extra conditions on sides and angles, quadrilaterals are grouped into special types:

• Trapezium: exactly one pair of opposite sides is parallel (AB ∥ CD).
• Kite: two pairs of adjacent sides are equal (AB = AD and CB = CD), but opposite sides are not equal.
• Parallelogram: both pairs of opposite sides are parallel (AB ∥ CD and AD ∥ BC).
• Rhombus: a parallelogram in which all four sides are equal.
• Rectangle: a parallelogram in which every angle is 90°.
• Square: a parallelogram that is both a rhombus and a rectangle — all sides equal and all angles 90°.

Every rectangle, rhombus and square is automatically a parallelogram, so it automatically has all the parallelogram properties (opposite sides equal, opposite angles equal, diagonals bisecting each other) plus its own extra property. A kite, however, is generally not a parallelogram, since its equal sides are adjacent rather than opposite.

Worked Example: A quadrilateral PQRS has all four sides equal, but its angles are 100°, 80°, 100° and 80° (not 90° each). What type of quadrilateral is PQRS?

All sides equal but angles are not 90° → PQRS is a parallelogram with all sides equal, which is exactly the definition of a rhombus (it is not a square, since the angles are not right angles).

Let ABCD be a parallelogram (AB ∥ CD and AD ∥ BC). Draw diagonal AC.

Theorem 1 — A diagonal divides a parallelogram into two congruent triangles. In ΔABC and ΔCDA: ∠BAC = ∠DCA (alternate angles, AB ∥ CD, transversal AC), ∠BCA = ∠DAC (alternate angles, AD ∥ BC, transversal AC), and AC = CA (common). By ASA, ΔABC ≅ ΔCDA.

Theorem 2 — Opposite sides are equal. Since ΔABC ≅ ΔCDA, their corresponding sides are equal: AB = CD and BC = AD.

Theorem 3 — Opposite angles are equal. From the same congruence, ∠B = ∠D. Since ∠BAC = ∠DCA and ∠DAC = ∠BCA, adding gives ∠A (= ∠BAC + ∠DAC) = ∠C (= ∠DCA + ∠BCA). So ∠A = ∠C and ∠B = ∠D.

Theorem 4 — Diagonals bisect each other. Let diagonals AC and BD meet at O. In ΔAOB and ΔCOD: AB = CD (opposite sides), ∠OAB = ∠OCD (alternate angles, AB ∥ CD), ∠OBA = ∠ODC (alternate angles). By ASA, ΔAOB ≅ ΔCOD, so OA = OC and OB = OD.

Worked Example: In parallelogram ABCD, ∠A = 72°. Find all the other angles.

Opposite angle: ∠C = ∠A = 72°. Adjacent angles are supplementary (co-interior angles on the transversal between two parallel sides): ∠B = 180° − 72° = 108°, and ∠D = ∠B = 108°.

The previous section proved properties of a parallelogram. The converse statements are equally important: they give quick tests to check whether a given quadrilateral is a parallelogram.

Condition 1: If both pairs of opposite sides of a quadrilateral are equal (AB = CD and AD = BC), then it is a parallelogram. Reason: Diagonal AC splits it into ΔABC and ΔCDA which become congruent by SSS (AB = CD, BC = AD, AC common), giving ∠BAC = ∠DCA, which are alternate angles ⇒ AB ∥ CD, and similarly AD ∥ BC.

Condition 2: If both pairs of opposite angles are equal (∠A = ∠C and ∠B = ∠D), then it is a parallelogram. Reason: By the angle sum property, ∠A + ∠B + ∠C + ∠D = 360°. Substituting ∠C = ∠A and ∠D = ∠B gives 2∠A + 2∠B = 360°, so ∠A + ∠B = 180°. Since ∠A and ∠B are co-interior angles on transversal AB, this makes AD ∥ BC. Similarly, ∠A + ∠D = 180°, which makes AB ∥ DC. Both pairs of opposite sides being parallel means ABCD is a parallelogram.

Condition 3: If the diagonals of a quadrilateral bisect each other (OA = OC and OB = OD), then it is a parallelogram. Reason: ΔAOB ≅ ΔCOD by SAS (OA = OC, OB = OD, vertically opposite ∠AOB = ∠COD), giving ∠OAB = ∠OCD, alternate angles ⇒ AB ∥ CD; similarly AD ∥ BC.

Condition 4: If one pair of opposite sides is both equal and parallel (AB ∥ CD and AB = CD), then the quadrilateral is a parallelogram. This is the most commonly used test in proofs.

Worked Example: In quadrilateral PQRS, PQ ∥ SR and PQ = SR = 6 cm. Is PQRS a parallelogram?

Since one pair of opposite sides (PQ and SR) is both equal and parallel, by Condition 4, PQRS is a parallelogram.

Mid-Point Theorem: In a triangle ABC, if D and E are the midpoints of sides AB and AC respectively, then DE is parallel to BC, and DE = ½ BC.

This is proved by extending DE to a point F such that DE = EF, joining CF, and showing that ΔAED ≅ ΔCEF (SAS, using AE = EC, ∠AED = ∠CEF vertically opposite, DE = EF). This gives AD = CF and AD ∥ CF. Since AD = DB (D is the midpoint), DB = CF as well, and DB ∥ CF, so BDFC is a parallelogram. Hence DF ∥ BC and DF = BC, which means DE (half of DF) is parallel to BC and DE = ½ BC.

Converse of the Mid-Point Theorem: The line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side. So, if D is the midpoint of AB and DE ∥ BC (with E on AC), then E must be the midpoint of AC.

Worked Example: In ΔABC, D and E are midpoints of AB and AC. If BC = 10 cm, find DE. Also, if ∠ADE = 65°, find ∠ABC.

By the mid-point theorem, DE = ½ BC = ½ × 10 cm = 5 cm.

Since DE ∥ BC, and AB is a transversal, ∠ADE and ∠ABC are corresponding angles, so ∠ABC = ∠ADE = 65°.