Given : A quadrilateral ABCD, in which diagonals AC and BD are equal and bisect each other at right angles,
To Prove : ABCD is a square.
Proof : Since ABCD is a quadrilateral whose diagonals bisect each other, so it is a parallelogram. Also, its diagonals bisect each other at right angles, therefore, ABCD is a rhombus.
⇒ AB = BC = CD = DA [Sides of a rhombus]
In ∆ABC and ∆BAD, we have
AB = AB [Common]
BC = AD [Sides of a rhombus]
AC = BD [Given]
∴ ∆ABC ≅ ∆BAD [SSS congruence]
∴ ∠ABC = ∠BAD [CPCT]
But, ∠ABC + ∠BAD = 180° [Consecutive interior angles]
∠ABC = ∠BAD = 90°
∠A = ∠B = ∠C = ∠D = 90° [Opposite angles of a ||gm]
⇒ ABCD is a rhombus whose angles are of 90° each.
Hence, ABCD is a square. Proved.
