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

(i) ABCD is a square 

(ii) diagonal BD bisects ∠B as well as ∠D.

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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

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