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Slovenian Mathematical Olympiad, 2008

Geometry Difficulty 5.4 AIME, harder Prove it Slovenia

In a quadrilateral $ABCD$ let $K$ be a point inside the triangle $ABD$ such that triangles $ABD$ and $KCD$ are similar. Prove that triangles $BCD$ and $AKD$ are similar as well.

Solution

Since triangles $ABD$ and $KCD$ are similar, we have $\angle ADB = \angle KDC$ and $\frac{|DA|}{|DB|} = \frac{|DK|}{|DC|}$ . We see that
$\angle ADK = \angle ADB - \angle BDK = \\ = \angle KDC - \angle BDK = \angle BDC$
and since $\frac{|DA|}{|DK|} = \frac{|DB|}{|DC|}$ , we conclude that triangles $ADK$ and $DBC$ are also similar (matching in one angle and the ratio of the two adjacent sides).

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