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Geometry Difficulty 6.3 National Olympiad Prove it Asia Pacific Mathematics Olympiad (APMO)

Let $ABCD$ be a quadrilateral such that all sides have equal length and angle $\angle ABC$ is $60$ degrees. Let $\ell$ be a line passing through $D$ and not intersecting the quadrilateral (except at $D$ ). Let $E$ and $F$ be the points of intersection of $\ell$ with $AB$ and $BC$ respectively. Let $M$ be the point of intersection of $CE$ and $AF$ .
Prove that $CA^{2} = CM \times CE$ .

Solution

Triangles $AED$ and $CDF$ are similar, because $AD \parallel CF$ and $AE \parallel CD$ . Thus, since $ABC$ and $ACD$ are equilateral triangles,
$\frac{AE}{CD} = \frac{AD}{CF} \Longleftrightarrow \frac{AE}{AC} = \frac{AC}{CF} .$
The last equality combined with
$\angle EAC = 180^{\circ} - \angle BAC = 120^{\circ} = \angle ACF$
shows that triangles $EAC$ and $ACF$ are also similar. Therefore $\angle CAM = \angle CAF = \angle AEC$ , which implies that line $AC$ is tangent to the circumcircle of $AME$ . By the power of a point, $CA^{2} = CM \cdot CE$ , and we are done.

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