In a quadrilateral ABCD, AB = AD and CB = CD. Prove that :

(i) AC bisects angle BAD.

(ii) AC is perpendicular bisector of BD.

(i) In △ ABC and △ ADC,

⇒ AB = AD (Given)

⇒ BC = CD (Given)

⇒ AC = AC (Common side)

∴ △ ABC ≅ △ ADC (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

⇒ ∠BAC = ∠DAC

∴ AC bisects ∠BAD.

Hence, proved that AC bisects angle BAD.

(ii) Since, AC bisects ∠BAD

∴ ∠BAO = ∠DAO

In △ AOB and △ AOD,

⇒ AB = AD (Given)

⇒ AO = AO (Common side)

⇒ ∠BAO = ∠DAO (Proved above)

∴ △ AOB ≅ △ AOD (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

⇒ ∠BOA = ∠DOA ……..(1)

From figure,

⇒ ∠BOA + ∠DOA = 180° (Linear pair)

⇒ ∠BOA + ∠BOA = 180° [From equation (1)]

⇒ 2∠BOA = 180°

⇒ ∠BOA = $\dfrac{180°}{2}$ = 90°.

∴ AC is perpendicular bisector of BD.

Hence, proved that AC is perpendicular bisector of BD.