Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Prove it Austria

Let $ABC$ be an acute triangle with $AB > AC$ . Let $D$ , $E$ and $F$ denote the feet of its altitudes on $BC$ , $AC$ and $AB$ , respectively. Let $S$ denote the intersection of lines $EF$ and $BC$ .
Prove that the circumcircles $k_1$ and $k_2$ of the two triangles $AEF$ and $DES$ touch in $E$ .

Solution

Figure 3: Problem 14

Let $t_1$ be the tangent line to $k_1$ in point $E$ and let $t_2$ be the tangent line to $k_2$ in point $E$ . The tangent-secant theorem applied to circle $k_1$ gives
$\angle(EF, t_1) = \angle FAE = \alpha$
with the usual notation for the angles in triangle $ABC$ .
The tangent-secant theorem applied to circle $k_2$ gives
$\angle(EF, t_2) = \angle SDE = \angle CDE = \alpha,$
where the last equality comes from the fact that $ABDE$ is a cyclic quadrilateral since all four vertices lie on the Thales circle with diameter $AB$ .
Therefore, $t_1$ and $t_2$ are parallel and they both contain the point $E$ . So, the two tangents are identical which implies that the circles touch in $E$ .

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