Question

# Prove that a cyclic parallelogram is rectangle.

Solution

## Given : $$ABCD$$ be a cyclic parallelogram.To prove :$$ABCD$$ is a rectangle.A rectangle is a parallelogram with one angle $$90^0$$So, we have to prove angle $$90^0$$Since $$ABCD$$ is a paralleogram$$\angle A = \angle C$$               (opposite angle of parallelogram are equal)......................(i)In cyclic parallelogram $$ABCD$$$$\angle A + \angle C=180^0$$      (Sum of opposite angles of a cyclic  quadrilateral is $$180^0$$)         $$\angle A + \angle A=180^0$$          from(i)$$2 \angle A =180^0$$$$\angle A = \dfrac{{{{180}^0}}}{2} = {90^0}$$Hence,$$\angle A = \angle C= 90^o$$Similarly we can prove other two angles are also equal to $$90^o$$Thus, a cyclic parallelogram is rectangle.Mathematics

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