Given that the vertices A, B, C of a quadrilateral ABCD lie on a circle.
Also $∠ A + ∠ C = 180^{o}$ , then prove that the vertex D also lie on the same circle.

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Solution

Let us consider a circle with center O passing through A, B, C

Given that $∠ A + ∠ C = 180$

Which implies Opposite angles of a quadrilateral. $A B C D$ are supplementary

$A B C D$ is a cyclic quadrilateral. $D$ also lies on the same circle.

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If the sides of a quadrilateral ABCD touch a circle, prove that :

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A B + B C + C D + D A < 2 (A C + B D)

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