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Saudi Arabian Mathematical Competitions

Geometry Difficulty 6.8 National Olympiad Prove it Saudi Arabia

Let $ABC$ be a non-isosceles triangle inscribed in a circle $(O)$ and $BE$ , $CF$ are two angle bisectors intersecting at $I$ with $E$ belonging to segment $AC$ and $F$ belonging to segment $AB$ . Suppose that $BE$ , $CF$ intersect $(O)$ at $M$ , $N$ respectively. The line $d_{1}$ passes through $M$ and is perpendicular to $BM$ , intersecting $(O)$ at the second point $P$ ; the line $d_{2}$ passes through $N$ and is perpendicular to $CN$ , intersecting $(O)$ at the second point $Q$ . Denote $H$ , $K$ as the midpoints of $MP$ and $NQ$ respectively.

1. Prove that triangles $IEF$ and $OKH$ are similar.

2. Suppose that $S$ is the intersection of the two lines $d_{1}$ and $d_{2}$ . Prove that $SO$ is perpendicular to $EF$ .

Solution

1) Denote the projections of $I$ on $AC$ , $AB$ are $X$ , $Y$ respectively.
Note that $B$ , $O$ , $P$ are collinear because $\angle BMP = 90^\circ$ . Because $O$ , $H$ are the midpoints of $PB$ , $PM$ respectively, we have $OH \parallel BM$ and $OH = \frac{1}{2} BM$ .
Similarly, $OK \parallel CN$ and $OK = \frac{1}{2} CN$ . Hence, $\angle HOK = \angle EIF$ . On the other hand,
$\begin{aligned} & \frac{OH}{OK} = \frac{BM}{CN} = \frac{\sin \left(A + \frac{B}{2}\right)}{\sin \left(A + \frac{C}{2}\right)} \\ & \frac{IF}{IE} = \frac{IY}{\sin \angle IFY} : \frac{IX}{\sin \angle IEX} = \frac{\sin \angle IEX}{\sin \angle IFY} = \frac{\sin \left(A + \frac{B}{2}\right)}{\sin \left(A + \frac{C}{2}\right)} \end{aligned}$
So $\frac{OH}{OK} = \frac{IF}{IE}$ implies that $\triangle OHK \sim \triangle IFE$ .

2) It is easy to see that
$OK \cdot OT = R^2, \quad OH \cdot OU = R^2$
with $R$ the radius of $(O)$ . Then $OK \cdot OT = OH \cdot OU$ or $K$ , $T$ , $H$ , $O$ are cyclic.
From this, we can see $\angle IEF = \angle IKH = \angle IUT$ so $EF \parallel UT$ .
The antipole line of $T$ , $U$ passes through $S$ so the antipole line of $S$ is $TU$ , which leads to $TU \perp SO$ . From these results, we get $SO \perp EF$ .

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