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Mongolian Mathematical Olympiad

Geometry Difficulty 5.2 AIME, harder Prove it Mongolia

Let $M$ be the midpoint of the side $BC$ of acute triangle $ABC$ and $H$ be the orthocenter of $ABC$ . Prove that if $D$ is base of perpendicular dropped from the vertex $A$ to the line $HM$ , then the intersection point of bisectors of the angles $DBH$ , $DCH$ lies on the line $HM$ .

Solution

Let $BB'$ , $CC'$ be altitudes. Then points $D$ , $B'$ , $C'$ lie on the circle with diameter $AH$ . Since $\angle C'DM = \angle C'DH = \angle C'AH = \angle C'CM$ , points $D$ , $C'$ , $M$ , $C$ are cyclic. Thus we have $\triangle C'HM \sim \triangle DHC \Rightarrow \frac{C'M}{CD} = \frac{HM}{HC} \Rightarrow \frac{CD}{CH} = \frac{C'M}{HM}$ .

Similarly, we conclude that $\frac{BD}{BH} = \frac{B'M}{HM}$ . Since $C'M = B'M$ , we get $\frac{CD}{CH} = \frac{BD}{BH}$ . Bisector of the angle $\angle BDH$ is $BL$ from where follows $\frac{BD}{DH} = \frac{BL}{LH}$ and the fact that $BL'$ is bisector of the angle $\angle CDH$ implies $\frac{CD}{CH} = \frac{CL'}{L'H} \Rightarrow L = L'$ .

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