Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.6 National Olympiad Prove it Hong Kong

Let $\triangle ABC$ be a scalene triangle, and let $D$ and $E$ be points on sides $AB$ and $AC$ respectively such that the circumcircles of triangles $ACD$ and $ABE$ are tangent to $BC$ . Let $F$ be the intersection point of $BC$ and $DE$ . Prove that $AF$ is perpendicular to the Euler line of $ABC$ .
(Recall that the Euler line of a triangle is the line passing through its circumcentre and the orthocentre.)

Solution

Let $O$ and $H$ be the circumcentre and orthocentre of $\triangle ABC$ respectively. Let $X$ and $Y$ be the points on the extension of $AB$ and $AC$ such that $HA = HX = HY$ . The the centres of $(ABC)$ and $(AXY)$ are $O$ and $H$ respectively. It remains to show that $F$ lies on the radical axis of these circles.

Firstly, since $\angle CXB = \angle CXA = \angle XAC = \angle BAY = \angle AYB = \angle CYB$ , the points $C, B, X, Y$ are concyclic. Also, since $\angle CBE = \angle BAE = \angle CXB$ , the line $BE$ is tangent to $(CBXY)$ . Similarly, $CD$ is tangent to $(CBXY)$ .

Applying Pascal's theorem to the points $BBCCYX$ , we know that the points $BB \cap CY = E$ , $CC \cap XB = D$ and $BC \cap YX$ are collinear. As $BC \cap DE = F$ , this means $X, Y, F$ are collinear. Now, the powers of $F$ with respect to $(ABC)$ and $(AXY)$ are $FB \times FC$ and $FX \times FY$ respectively, which are equal since $C, B, X, Y$ are concyclic. Therefore, $F$ lies on the radical axis of these circles. Thus, $AF$ is the radical axis. This implies $AF \perp OH$ .

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