# ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠C. Show that : (i) ABCD is a square

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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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