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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.

1193351_1310445_ans_13432f17bbf0437da4570481a055705d.png

Mathematics

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