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National Math Olympiad in Slovenia

Geometry Difficulty 4.7 AIME Prove it Slovenia

In the parallelogram $ABCD$ we have $|AB| = |BD|$ . Let $K$ be a point on the line $AB$ different from $A$ , such that $|KD| = |AD|$ . Denote the reflection of the point $C$ over $K$ by $M$ , and the reflection of $B$ over $A$ by $N$ . Prove that $MDN$ is an isosceles triangle with the apex at $D$ .

Solution

Triangle $ADK$ is isosceles with the apex at $D$ . So, $\angle BKD = 180^\circ - \angle DKA = 180^\circ - \angle KAD = \angle CBK$ and $|DK| = |DA| = |BC|$ . The triangles $DKB$ and $CBK$ are congruent because they have a common side $KB$ , congruent angles $\angle BKD = \angle CBK$ and $|DK| = |BC|$ . Thus, $\angle DCK = \angle BKC = \angle DBK$ . Since $M$ and $N$ are reflections of $C$ and $B$ , we have $|CK| = |KM|$ and $|NA| = |AB|$ . So,
$|CM| = 2|CK| = 2|DB| = 2|AB| = |NB|.$
This implies that the triangles $CDM$ and $BDN$ are also congruent since $|DC| = |DB|$ , $|CM| = |NB|$ and $\angle DCM = \angle DBN$ . Thus, $|DN| = |DM|$ and the triangle $DMN$ is isosceles with the apex at $D$ .

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