A Very Convex Quadrilateral | Gonitzoggo

A Very Convex Quadrilateral

$ABCD$ is a convex quadrilateral such that $\angle ABD = \angle DBC, AD= CD$ and $AB \neq BC.$ Prove that $ABCD$ is cyclic.

(A convex quadrilateral is a quadrilateral having all of its interior angles measuring less than $180^{\circ}$, A convex quadrilateral is cyclic if and only if the sum of its two opposite angles is $180^{\circ}$.)

Proof Based Problems

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Solution

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Let $D'$ be the intersection of line $BD$ and circle $ABC$.

$ABCD'$ is cyclic. So,
\[\angle D'AC=\angle D'BC=\angle DBC=\angle ABD=\angle ABD'=\angle ACD'\]

$\therefore D'$ lies on the perpendicular bisector of $AC$ and it also lies on $BD$.

$AD=CD\implies D$ also lies on the perpendicular bisector of $AC$.

Two lines can intersect at atmost one point.
\[\therefore D=D'\]
So, $D$ is on circle $ABC$ or $ABCD$ is cyclic.

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