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

Since, opposite angles of a parallelogram are equal.

∴ ∠ABC = ∠ADC = x (let)

Given,

BD bisects ∠B.

∴ ∠ABD = ∠CBD = $\dfrac{1}{2} ∠ABC = \dfrac{x}{2}$

BD bisects ∠D.

∴ ∠ADB = ∠BDC = $\dfrac{1}{2} ∠ADC = \dfrac{x}{2}$

⇒ ∠ABD = ∠ADB and ∠CBD = ∠CDB

In △ ABD,

⇒ ∠ABD = ∠ADB

∴ AB = AD (Sides opposite to equal angles are equal) ……(1)

In △ CBD,

⇒ ∠CBD = ∠CDB

∴ CD = BC (Sides opposite to equal angles are equal) ………(2)

As, ABCD is a parallelogram.

Thus, opposite sides are equal.

∴ AB = CD ………(3)

∴ AD = BC ……….(4)

From equations (1), (2), (3) and (4), we get :

⇒ AB = BC = CD = AD.

Since, all sides of quadrilateral ABCD are equal.

Hence, proved that ABCD is a rhombus.