Special Types of Quadrilaterals

168. The Rectangle's Diagonals · equal length, mutual bisectors

The diagonals of a rectangle are always equal in length and bisect each other.

In a rectangle, the diagonals are equal in length AND bisect each other. They are not perpendicular (unless the rectangle is also a square). The two diagonals split the rectangle into four congruent isosceles triangles.

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Selina ICSE: Special Types of Quadrilaterals

What this lesson covers

Try to break it

Drag W to widen and H to make it taller. The diagonals AC and BD always come out the same length and always meet at O — the midpoint of both. Try to drag W or H to make AC ≠ BD; you can't. That's the rectangle diagonal theorem.

How you build it

Construct a rectangle with diagonals.

Mark point O — the centre where both diagonals will bisect each other.
Draw a circle centred at O — this is the circumscribed circle. Every vertex of the rectangle will lie on it.
Mark point A anywhere on the circle — the first vertex.
Draw a line through O and A — it extends to the far side of the circle, where diagonal AC will end.
Mark point C where the line exits the circle on the far side of O from A. O is now the midpoint of diagonal AC.
Mark point B anywhere on the circle — the third vertex. Choose a position well away from A and C.
Draw a line through O and B — it extends to the far side of the circle, where diagonal BD will end.
Mark point D where the line exits the circle on the far side of O from B. O bisects BD, and BD = AC (both are diameters). ABCD is a rectangle.
Draw diagonal BD — both diagonals are equal (AC = BD = 2r) and bisect each other at O.
Draw side AB.
Draw side BC.
Draw side CD.
Draw side DA to complete rectangle ABCD.

The proof, step by step

Prove that the diagonals of a rectangle are equal and bisect each other.

In ΔABC and ΔBAD: AB = AB (common), BC = AD (opposite sides of rectangle), ∠ABC = ∠BAD = 90°. ∴ ΔABC ≅ ΔBAD [SAS].
From congruence, AC = BD (CPCT). Hence, diagonals are equal.
ABCD is a rectangle, so it's a parallelogram. AD || BC. In ΔAOD and ΔCOB: AD = BC, ∠ADO = ∠CBO (alt. angles), ∠DAO = ∠BCO (alt. angles). ∴ ΔAOD ≅ ΔCOB [ASA].
From congruence, OA = OC and OB = OD (CPCT). Hence, diagonals bisect each other.

Worked example

In rectangle ABCD, diagonals AC and BD intersect at O. If AC = 14 cm, what is the length of OB?

Diagonals of a rectangle are equal, so BD = AC = 14 cm. They also bisect each other, so O is the midpoint of BD. Thus, OB = BD/2 = 7 cm.

7 cm — correct
14 cm
28 cm
10 cm