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China Southeastern Mathematical Olympiad

Geometry Difficulty 4.9 AIME Prove it China

As shown in the figure, $BC$ and $AC$ are tangent to the inscribed circle $I$ of $\triangle ABC$ at $M$ and $N$ . $E$ and $F$ are the midpoints of $AB$ , $AC$ respectively. $EF$ intersects $BI$ at $D$ . Prove that $M$ , $N$ , $D$ are collinear.

Solution

Join $AD$ , and it is obvious that $\angle ADB = 90^\circ$ . Then join $AI$ and $DM$ . Suppose $DM$ intersects $AC$ at $G$ . Since $\angle ABI = \angle DBM$ , we obtain $\frac{AB}{BD} = \frac{BI}{BM}$ . Hence, $\triangle ABI \sim \triangle DBM$ , and
$\angle DMB = \angle AIB = 90^\circ + \frac{1}{2} \angle ACB.$
Join $IG$ , $IC$ , $IM$ , then
$\angle IMG = \angle DMB - 90^\circ = \frac{1}{2} \angle ACB = \angle GCI.$
Therefore, $I$ , $M$ , $C$ , $G$ are concyclic, and $IG \perp AC$ .
Consequently, since $G$ and $N$ represent the same point, we conclude that $M$ , $N$ , $D$ are collinear.

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