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Turkish Mathematical Olympiad

Geometry Difficulty 6.4 National Olympiad Prove it Turkey

Let $D$ be a point on the side $[BC]$ of the triangle $ABC$ different from the vertices and $E$ be the midpoint of $[CD]$ . The line perpendicular to $BC$ at $E$ intersects the side $[AC]$ at point $F$ satisfying $AF \cdot BC = AC \cdot EC$ . Let $G$ be the second point where the circumcircle of the triangle $ADC$ intersects the side $[AB]$ . Prove that the tangent line of the circumcircle of the triangle $AGF$ at $F$ is also tangent to the circumcircle of the triangle $BGE$ .

Solution

We will show that $EF$ is a common tangent of the circumcircles of $AGF$ and $BGE$ . Note that it is enough to show that $\angle GBE = \angle GEF$ and $\angle GAF = \angle GFE$ .

Let $H$ be the point of intersection of $AB$ and the line passing through $F$ parallel to $BC$ .

Then $AHF \sim ABC$ and $\frac{AF}{AC} = \frac{HF}{BC}$ . On the other hand, it is given that $\frac{AF}{AC} = \frac{EC}{BC}$ . Therefore, $HF = EC = ED$ and hence $HFCE$ is a parallelogram and $HFDE$ is a rectangle.

Since $ACDG$ is a cyclic quadrilateral and $FC \parallel HE$ , we have $\angle BGD = \angle ACB = \angle HED$ . Then $H, E, D, G$ are cyclic. Consequently, $H, G, D, E, F$ are on the circle of diameter $[HE]$ . Then $\angle BGE = 90^\circ$ and $\angle GBE = 90^\circ - \angle GED = \angle GEF$ . Since $AGDC$ and $GFED$ are cyclic quadrilaterals $\angle BAC = 180^\circ - \angle GDE = \angle GFE$ .

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