Question

The diagonal BD of a parallelogram ABCD bisects angles B and D. Prove that ABCD is a rhombus.

Answer 1

Given: ABCD is a parallelogram where the diagonal BD bisects
parallelogram  ABCD at angle B and D

To Prove:  ABCD is a rhombus

Proof: Let us draw a parallelogram  ABCD where the diagonal BD bisects the parallelogram at an angle B and D.

Construction: Let us join AC as a diagonal of the parallelogram ABCD

Since ABCD  is a parallelogram in which diagonal BD bisects ∠B and ∠D

BD bisect ∠B,

∠ABD = ∠CBD = `1/2`∠ABC

BD bisect ∠D,

∠ADB = ∠CDB = `1/2`∠ABC

∴ ∠ABC = ∠ADC    ...(Opposite angles of parallelogram are equal)

∴ `1/2`∠ABC = `1/2` ∠ADC

∠ABD = ∠ADB and ∠CBD = ∠CDB

∴ AD = AB and CD = BC

As ABCD is a parallelogram, opposite sides are equal

AB = CD and AD = BC

AB = BC = CD = AD

∴ ABCD is a rhombus.