Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Croatia

An isosceles triangle $ABC$ with $|AB| = |AC|$ is given. Tangents to its circumcircle at points $A$ and $C$ intersect in $D$ . If $\angle DBC = 30^\circ$ , prove that triangle $ABC$ is equilateral.

Solution

Let $BCE$ and $ECF$ be the equilateral triangles such that the points $A$ and $E$ are at the same side of the line $BC$ , and $B$ and $F$ at different sides of the line $CE$ . We will assume that $E$ and $A$ are different points, otherwise the statement holds.

Since $\angle FBC = 30^\circ$ , and then $\angle FBC = \angle DBC$ , we conclude that $B$ , $F$ and $D$ are collinear. Triangles $DFC$ and $DFE$ are congruent (SAS), so $|DE| = |DC| = |DA|$ . Since $AE$ is perpendicular to $BC$ and then also to $AD$ , from the right triangle $DAE$ we see that $|DE| > |DA|$ , which is a contradiction.

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