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If the diagonals of a parallelogram are equal, then show that it is a rectangle

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Given : ABCD is a parallelogram in which AC = BD. 

To Prove : ABCD is a rectangle. 

Proof : In ∆ABC and ∆ABD 

AB = AB [Common] 

BC = AD [Opposite sides of a parallelogram] 

AC = BD [Given] 

∴ ∆ABC ≅ ∆BAD [SSS congruence] 

∠ABC = ∠BAD …(i) [CPCT] 

Since, ABCD is a parallelogram, thus, 

∠ABC + ∠BAD = 180° …(ii) [Consecutive interior angles] 

∠ABC + ∠ABC = 180° 

∴ 2∠ABC = 180° [From (i) and (ii)] 

⇒ ∠ABC = ∠BAD = 90° 

This shows that ABCD is a parallelogram one of whose angle is 90°. Hence, ABCD is a rectangle. Proved.

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