Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National olympiad Prove it Greece

Let $A B \Gamma$ be an acute angled triangle with circumcircle $\omega$ . A circle $\gamma$ with center $A$ intersects the arc $AB$ of the circle $\omega$ , not containing $\Gamma$ , at point $\Delta$ and the arc $A\Gamma$ , not containing $B$ , at point $E$ . We suppose that the point of intersection $K$ of the lines $BE$ and $\Gamma\Delta$ belongs to the $\gamma$ . Prove that the line $AK$ is perpendicular to the line $B\Gamma$ .

Solution

From the relationship of a central angle and an inscribed angle that go on the same arc $\Delta K$ of the circle $\gamma$ , we have:
$\angle A\Delta K = 2 \cdot \angle E\Delta K \quad (1)$
Also we have the equality of inscribed angles
$\angle E\Delta K = \angle AB\Delta \quad (2)$
From relations (1) and (2) it follows that $\angle A\Delta K = 2 \cdot \angle AB\Delta$ , that is, $AB$ is the bisector of the angle $\angle A\Delta K$ .

Figure 1

Since $A\Delta = AK$ , (radii of the circle $\gamma$ ), the triangle $\Delta AK$ is isosceles and hence $AB \perp \Delta K$ , that is $AB \perp \Gamma\Delta$ . Similarly we get that $A\Gamma \perp BE$ . Hence $K$ is the orthocenter of the triangle $AB\Gamma$ , and therefore $AK \perp B\Gamma$ .

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