65th Czech and Slovak Mathematical Olympiad

Question

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 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) FIGURE_0

Accepted Answer

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. FIGURE_0 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.