Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 5.6 AIME, harder Prove it Czech Republic

Let $ABC$ be a right-angled triangle with a hypotenuse $AB$ and longer leg $BC$ . Let $D$ be a foot of an altitude from the vertex $C$ . Circle $k$ with the center $D$ and the radius $CD$ intersects the leg $BC$ in a point $Q$ and line $AB$ in points $E$ and $F$ ( $E \neq F$ ), where $F$ is a point on the hypotenuse $AB$ . Segment $QE$ intersects the leg $AC$ in a point $P$ . Prove that $PE = QF$ . (Jaroslav Švrček)

Solution

The circle $k$ is the Thales' circle with the diameter $EF$ and the center $D$ . A triangle $EFC$ is the isosceles right-angled triangle, so $EC = EF$ . We will show that triangles $EPC$ and $FQC$ are congruent, which will prove the statement of the problem.

Fig. 2

Angles $CEQ$ and $CFQ$ are congruent as they are inscribed angles subtended by the chord $CQ$ of the circle $k$ . Both angles $ECF$ and $ACB$ are congruent (right angles), so their remaining non-overlapping parts (angles $ECF = ECP$ and $ACB = FCQ$ ) are also congruent. This proves, that triangles $EPC$ and $FQC$ are congruent by $A$ - $S$ - $A$ .

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