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Saudi Arabian Mathematical Competitions

Geometry Difficulty 6.1 National Olympiad Prove it Saudi Arabia

Let $ABC$ be a triangle inscribed in the circle $(O)$ . The bisector of $\angle BAC$ cuts the circle $(O)$ again at $D$ . Let $DE$ be the diameter of $(O)$ . Let $G$ be a point on $\operatorname{arc} AB$ which does not contain $C$ . The lines $GD$ and $BC$ intersect at $F$ . Let $H$ be a point on the line $AG$ such that $FG \parallel AE$ . Prove that the circumcircle of triangle $HAB$ passes through the orthocenter of triangle $HAC$ .

Solution

Let $HF$ cut $DE$ at $P$ . We have $\angle HGD = \angle E = \angle HPD$ so $HGPD$ is cyclic, we deduce $FH \cdot FP = FG \cdot FD = FB \cdot FC$ so $HBPC$ is cyclic.

Easily seen $DE$ is perpendicular bisector of $BC$ so $PB = PC$ . Hence $HP$ is bisector of $\angle BHC$ . From this,
$\begin{aligned} \angle HBA + \angle HCA & = 180^\circ - \angle BHA - \angle BAH + 180^\circ - \angle CHA - \angle ACH \\ & = 360^\circ - (\angle AHB + \angle AHC) - (\angle HAB + \angle HAC) \\ & = 360^\circ - 2\angle AHF - 2\angle HAD \\ & = 360^\circ - 2\angle GDE - 2(\angle GAE - 90^\circ) = 180^\circ . \end{aligned}$
Now let $K$ be orthocenter of triangle $HAC$ then
$\angle AKH = 180^\circ - \angle ACH = \angle ABH$
so $K$ lies on $(HAB)$ . $\square$

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