Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it JBMO

Problem:

Around the triangle $A B C$ the circle is circumscribed, and at the vertex $C$ tangent $t$ to this circle is drawn. The line $p$ which is parallel to this tangent intersects the lines $B C$ and $A C$ at the points $D$ and $E$ , respectively. Prove that the points $A, B, D, E$ belong to the same circle.

Solution

Solution:

Let $O$ be the center of a circumscribed circle $k$ of the triangle $A B C$ , and let $F$ and $G$ be the points of intersection of the line $C O$ with the line $p$ and the circle $k$ , respectively (see Figure).

From $p \parallel t$ it follows that $p \perp C O$ .

Furthermore, $\angle A B C = \angle A G C$ , because these angles are peripheral over the same chord.

The quadrilateral $A G F E$ has two right angles at the vertices $A$ and $F$ , and hence, $\angle A E D + \angle A B D = \angle A E F + \angle A G F = 180^{\circ}$ .

Hence, the quadrilateral $A B D E$ is cyclic, as asserted.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.