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NMO Selection Tests for JBMO

Geometry Difficulty 6.8 National Olympiad Prove it Romania

In the exterior of the acute-angled triangle $ABC$ , we construct the isosceles triangles $DAB$ and $EAC$ with bases $AB$ and $AC$ , respectively, such that $\angle DBC = \angle ECB = 90^\circ$ . Let $M$ and $N$ be the reflections of $A$ with respect to $D$ and $E$ , respectively. Prove that the line $MN$ passes through the orthocenter of the triangle $ABC$ .

Solution

Since $AH \perp BC$ and $CP \perp BC$ , we deduce that $AH \parallel CP$ . Therefore $AHCP$ is a parallelogram, and $CP = AH$ .

Similarly, we deduce that $AHFC'$ is a parallelogram, hence $C'F = AH = CP$ . Since $C'E = CE$ , we obtain that $EP = EF$ .

Consequently, $D$ and $E$ are the midpoints of the bases of the trapezoid $BB'PF$ , and $Q$ is the intersection point of the lines $B'P$ and $BF$ , therefore $D, E, Q$ are collinear. Since $D$ and $I$ are the midpoints of the bases of the trapezoid $AHBB'$ , and $Q$ is the intersection point of the lines $B'A$ and $BH$ , it follows that $D, I, Q$ are also collinear, consequently $I$ lies on the line $DE$ .

$DI$ and $QI$ are midlines in the triangles $AMH$ and $ANH$ , respectively, hence $DI \parallel MH$ and $IE \parallel NH$ , therefore $H \in MN$ .

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