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Geometry Difficulty 5.6 AIME, harder Prove it South Africa

Let $CH$ be the altitude of an acute angled triangle $ABC$ , and let $O$ be the centre of its circumcircle. If $T$ is the foot of the perpendicular drawn from vertex $C$ to the line $AO$ , prove that the line $TH$ bisects the side $BC$ .

Solution

Let $TH$ cut $BC$ at $P$ . We want to show that $P$ bisects $BC$ . Since $\angle AHC = \angle ATC = 90^\circ$ , the points $A$ , $H$ , $T$ and $C$ lie on the same circle.

Now $\angle PHC = \angle TAC = \frac{1}{2}(180^\circ - \angle AOC) = 90^\circ - \angle ABC = 90^\circ - \angle HBC = \angle BCH = \angle PCH$ , so triangle $PCH$ is isosceles. In addition, $\angle PBH = 90^\circ - \angle BCH = 90^\circ - \angle PHC = \angle PHB$ , so the triangle $PBH$ is isosceles as well. Therefore $PC = PH = PB$ and the claim is proved.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.