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China Hong Kong Mathematical Olympiad

Geometry Difficulty 6.7 National Olympiad Prove it Hong Kong

Let $ABC$ be a triangle. Let $D$ and $E$ be respectively points on the segments $AB$ and $AC$ , and such that $DE \parallel BC$ . Let $M$ be the midpoint of $BC$ . Let $P$ be a point such that $DB = DP$ , $EC = EP$ and such that the open segments (segments excluding the endpoints) $AP$ and $BC$ intersect. Suppose $\angle BPD = \angle CME$ . Show that $\angle CPE = \angle BMD$ .

Solution

Note that $\angle BPD = \angle CME = \angle DEM$ . So we can find a point $F$ on the extension of $EM$ such that $\triangle DPB \sim \triangle DEF$ . By spiral similarity and since

$DB = DP$ , we have $\triangle DFB \cong \triangle DEP$ . As $EC = EP = FB$ and $CM = BM$ , by the sine law, either $\angle FEC = \angle EFB$ or $\angle FEC + \angle EFB = 180^\circ$ .
If $\angle FEC + \angle EFB = 180^\circ$ , then
$\begin{align*} \angle AED &= 180^\circ - \angle FEC - \angle DEM \\ &= \angle EFB - \angle DEM \\ &= \angle DFB + \angle EFD - \angle DEM \\ &= \angle DEP. \end{align*}$
Together with $\angle AEP = 2\angle ECP$ , we easily find that $DE \parallel PC$ , contradicting the fact that $P$ does not lie on $BC$ .

Hence, we must have $\angle FEC = \angle EFB$ . This yields $EC \parallel BF$ . Therefore, we find that $EM = FM$ . Since $DE = DF$ , we have $DM \perp EF$ . Now, by swapping the roles of $(B, D)$ and $(C, E)$ , and reversing the above arguments, we see that $\angle CPE = \angle BMD$ as desired.

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