Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 AIME, harder Prove it

2. In a right triangle $ABC$ , let $P$ be the foot of the altitude from vertex $C$ to the hypotenuse $AB$ . The intersection of segment $AB$ with the line passing through vertex $C$ and the center of the incircle of triangle $PBC$ is denoted as $D$ . Prove that segments $AD$ and $AC$ are congruent.

Solution

2. In a right-angled triangle $ABC$ with hypotenuse $AB$ , for the angles $\alpha, \beta$ at vertices $A, B$ , it holds that $\alpha+\beta=90^{\circ}$ . Therefore, $|\Varangle ACP|=90^{\circ}-\alpha=\beta$ and $|\Varangle BCD|=|\nless DCP|=$ $=\frac{1}{2}\left(90^{\circ}-\beta\right)=\frac{1}{2} \alpha$ , since line $CD$ is the angle bisector of $\angle BCP$ (Fig. 1). For the exterior angle $\angle ADC$ of triangle $BCD$ , it clearly holds that $|\nless ADC|=|\nless DBC|+|\nless BCD|=\beta+\frac{1}{2} \alpha=|\nless DCA|$ .

We have found that triangle $ADC$ has equal internal angles at vertices $C, D$ , so it is isosceles, and thus $|AD|=|AC|$ .

Fig. 1

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.