Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it China

In $\triangle ABC$ , $AB = AC$ , the inscribed circle $I$ touches $BC$ , $CA$ , $AB$ at points $D$ , $E$ and $F$ respectively. $P$ is a point on arc $\widehat{EF}$ (not containing $D$ ). Line $BP$ intersects the circle $I$ at another point $Q$ , and lines $EP$ , $EQ$ meet line $BC$ at $M$ , $N$ respectively. Prove that

(1) $P$ , $F$ , $B$ , $M$ are concyclic;

$(2) \frac{EM}{EN} = \frac{BD}{BP}.$

Solution

Proof

(1) From the given condition, $EF \parallel BC$ , so
$\begin{aligned} \angle ABC &= \angle AFE = \angle AFP + \angle PFE \\ &= \angle PEF + \angle PFE = 180^\circ - \angle FPE, \end{aligned}$
and thus $P$ , $F$ , $B$ , $M$ are concyclic.

(2) By the sine law, $EF \parallel BC$ and the fact that $P$ , $F$ , $B$ , $M$ are concyclic, we have
$\frac{EM}{EN} = \frac{\sin\angle ENM}{\sin\angle EMN} = \frac{\sin\angle FEN}{\sin(\pi - \angle PFB)} = \frac{\sin\angle FPB}{\sin\angle PFB} = \frac{BF}{BP}.$
Together with $BF = BD$ , the proposition is proven.

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