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Problem 768

AMC 12 late, AIME early

Geometry Difficulty 4.2 Prove it Japan Junior Mathematical Olympiad · Japan

Let $H$ be the orthocenter of an acute triangle $ABC$ , and let $D$ be the intersection of the two lines $AH$ , $BC$ . Let $E$ be the point of intersection of the circumcircle to the triangle $ABD$ and the line $CH$ , lying outside of the triangle $ABC$ . And let $F$ be the point of intersection of the circumcircle to the triangle $ACD$ and the line $BH$ , lying outside of the triangle $ABC$ . Show that the two line segments $AE$ and $AF$ have the same length.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $K$ , $L$ be the feet of the perpendicular lines drawn from $B$ to the side $CA$ and from $C$ to the side $AB$ , respectively. Since the line segment $AB$ is a diameter of the circumcircle to the triangle $ABD$ , $\angle AEB = 90^\circ$ . We see that the triangles $AEB$ and $ALE$ are similar, since they have the angles of same magnitudes. Therefore, we have $AE : AL = AB : AE$ , and we get $AE^2 = AB \cdot AL$ .

Similarly, we obtain $AF^2 = AC \cdot AK$ from the similarity of the triangles $AFC$ and $AKF$ . Since $\angle BKC = \angle BLC = 90^\circ$ , we see that the points $B$ , $C$ , $K$ , $L$ lie on the circumference of a same circle. Using the well-known theorem on the power of a point with respect to a circle, we then obtain $AB \cdot AL = AC \cdot AK$ , and therefore, we obtain $AE^2 = AF^2$ , which shows that $AE = AF$ .

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