**Given : **ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

**To Prove : **

(i) ABCD is a square.

(ii) Diagonal BD bisects ∠B as

well as ∠D.

**Proof :** (i) In ∆ABC and ∆ADC, we have

∠BAC = ∠DAC [Given]

∠BCA = ∠DCA [Given]

AC = AC

∴ ∆ABC ≅ ∆ADC [ASA congruence]

∴ AB = AD and CB = CD [CPCT]

But, in a rectangle opposite sides are equal,

i.e., AB = DC and BC = AD

∴ AB = BC = CD = DA

Hence, ABCD is a square **Proved.**

(ii) In ∆ABD and ∆CDB, we have

AD = CD

AB = CD [Sides of a square]

BD = BD [Common]

∴ ∆ABD ≅ ∆CBD [SSS congruence]

So, ∠ABD = ∠CBD

∠ADB = ∠CDB

Hence, diagonal BD bisects ∠B as well as ∠D **Proved**