Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 5.2 AIME, harder Prove it Italy

Let $ABC$ be a right triangle with the right angle at $A$ , with $AB > AC$ ; let $AH$ be the altitude relative to the hypotenuse. On the line $BC$ take $D$ such that $H$ is the midpoint of $BD$ ; let $E$ be the foot of the perpendicular drawn from $C$ to $AD$ . Prove that $EH = AH$ .

Solution

Solution:

The triangle $ABD$ is isosceles on $BD$ , because $AH$ is both an altitude and a median. $AH$ is therefore also the bisector of the angle $B\widehat{A}D$ , and hence the angles $D\widehat{A}H$ , $B\widehat{A}H$ are equal.

The angles $A\widehat{E}C$ , $A\widehat{H}C$ are right angles by construction; therefore $E$ and $H$ belong to the circle $\gamma$ having $AC$ as diameter.

$D\widehat{A}H$ and $B\widehat{A}H$ are inscribed angles with respect to $\gamma$ ; $D\widehat{A}H$ subtends the arc $EH$ , $BAH$ subtends the arc $AH$ (the latter is in the "limiting position", since the side $AB$ is tangent to $\gamma$ at $A$ ).

Therefore, the arcs $EH$ , $AH$ of $\gamma$ are equal, and hence the chords $EH$ , $AH$ are also equal, as was to be shown.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.