Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Prove it Romania

Consider a convex quadrilateral $ABCD$ having $\angle BCD = 120^\circ$ , $\angle CBA = 45^\circ$ , $\angle CBD = 15^\circ$ and $\angle CAB = 90^\circ$ . Show that $AB = AD$ .

Solution

Consider the mirror image $E$ of point $C$ with respect to $A$ . Since $\angle CDB = \angle CEB = 45^\circ$ , the quadrilateral $BCDE$ is cyclic, and $CE$ is a diameter of its circumcircle. Hence $A$ is its circumcenter, implying $AB = AD$ .

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