Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 4.9 AIME Prove it Ukraine

In the plane let be given a circle with a chord $AB$ and a point $P$ on this chord such that $AP = 2PB$ . The chord $DE$ is normal to the chord $AB$ and passing through the point $P$ . Prove that the midpoint of the segment $AP$ is the orthocenter of the triangle $AED$ .

Solution

Let $M$ be a midpoint of the segment $AP$ . Let us the line $EM$ intersects the segment $AD$ at a point $Q$ . $MPE$ and $BPE$ are the congruent triangles because both of them are right and both of them have the equal legs. Then $\angle QAM = \angle DAB = \angle DEB = \angle PEB = \angle PEM$ (Fig.9).

Since $\angle QMA = \angle EMP$ then triangles $QMA$ and $EMP$ have in twos equal angles. Consequently both of them have equal third angles, then $\angle AQM = \angle MPE = 90^\circ$ , then $EQ$ and $AP$ are the altitudes of the triangle $ADE$ which proves that $M$ is the orthocenter of that triangle. That completes the proof.

Fig.9

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