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Geometry Difficulty 4.5 AIME Prove it South Africa

From a point $P$ , draw tangents $PB$ and $PT$ to a circle. Let $A$ be the point on the circle such that $AB$ is a diameter, and let $H$ be the foot of the perpendicular from $T$ onto $AB$ . Prove that $AP$ bisects $TH$ .

Solution

Construct $P'$ on $AT$ such that $OP' \perp AB$ . Then $OP' \parallel HT$ . Since $BT$ is perpendicular to both $OP$ and $AT$ , we have $OP \parallel AT$ , and by the midpoint theorem $OP'$ bisects $AP$ .

Therefore $OAP'P$ is a parallelogram. Hence $OP'$ is bisected by $AP$ , and since triangles $HAT$ and $OAP'$ are similar, $AP$ bisects $HT$ as required.

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