Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it Ireland

Let $ABC$ be an isosceles triangle with $|AB| = |AC|$ . The points $D, E$ and $F$ are on the sides $BC, CA$ and $AB$ , respectively, such that $\angle FDE = \angle ABC$ and $FE$ is not parallel to $BC$ . Prove that $BC$ is tangent to the circumcircle of $\triangle DEF$ if and only if $D$ is the midpoint of $BC$ .

Solution

Assume first that $BC$ is tangent to the circumcircle of $\triangle DEF$ . Then $\angle EDC = \angle DFE$ . By assumption $\angle FDE = \angle ABC = \angle ACB$ , thus the two triangles $DEC$ and $DEF$ are similar. Hence, $\frac{|BD|}{|FD|} = \frac{|DE|}{|FE|}$ . Similarly we obtain that $\triangle FBD$ and $\triangle DEF$ are similar and so $\frac{|CD|}{|DE|} = \frac{|FD|}{|FE|}$ . We now easily see that $|BD| = |CD|$ .

On the other hand, let $D$ be the midpoint of $BC$ and let the circumcircle of $\triangle DEF$ meet $AB$ at $H$ again (as shown in the diagram).

Then $\angle EDF = \angle EHF$ hence $\angle EHF = \angle ABC$ and so $HE$ is parallel to $BC$ . Because $AD$ is perpendicular to $BC$ and bisects $BC$ , $AD$ is perpendicular to $HE$ and bisects $HE$ as well. Therefore, the centre of the circumcircle of $\triangle DEF$ , which also is the circumcircle of $\triangle HEF$ , lies on the line $AD$ . This implies that $BC$ is tangent to the circumcircle of $\triangle DEF$ .

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