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Estonian Mathematical Olympiad

Geometry Difficulty 5.9 AIME, harder Prove it Estonia

An acute-angled triangle $ABC$ with $AC > AB$ is given. The perpendicular bisector of side $BC$ intersects the lines $AC$ and $AB$ at points $D$ and $E$ , respectively. The circle with diameter $DE$ intersects the lines $AC$ and $AB$ at points $K$ and $L$ , respectively ( $K \neq D, L \neq E$ ). Let $M$ be the midpoint of side $BC$ . Prove that the points $K, L$ , and $M$ are collinear.

Solution

Since the points $E, K, L$ , and $D$ are concyclic (Fig. 28), it follows that $\angle CKL = \angle DKL = \angle DEL = \angle MEB$ . From the problem conditions, $\angle CME = 90^\circ$ , and by Thales' theorem, $\angle CKE = \angle DKE = 90^\circ$ . Therefore, the points $C, M, K, E$ are also concyclic. Consequently, $\angle CKM = \angle CEM$ . Since point $E$ lies on the perpendicular bisector of side $BC$ , the triangles $BEM$ and $CEM$ are congruent, thus $\angle MEB = \angle CEM$ . From the previous result, $\angle CKL = \angle CKM$ . Since the points $L$ and $M$ lie on the same side of line $CK$ , the points $K, L$ , and $M$ are collinear.

Fig. 28

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