Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.3 National olympiad Prove it

We are given an acute triangle $A B C$ . Let $D$ be the point on its circumcircle such that $A D$ is a diameter. Suppose that points $K$ and $L$ lie on segments $A B$ and $A C$ , respectively, and that $D K$ and $D L$ are tangent to circle $A K L$ .

Show that line $K L$ passes through the orthocentre of $A B C$ .
The altitudes of a triangle meet at its orthocentre.
!

Figure 1: Diagram to solution 1

Solution

Let $M$ be the midpoint of $K L$ . We will prove that $M$ is the orthocentre of $A B C$ . Since $D K$ and $D L$ are tangent to the same circle, $|D K|=|D L|$ and hence $D M \perp K L$ . The theorem of Thales in circle $A B C$ also gives $D B \perp B A$ and $D C \perp C A$ . The right angles then give that quadrilaterals $B D M K$ and $D M L C$ are cyclic.

If $\angle B A C=\alpha$ , then clearly $\angle D K M=\angle M L D=\alpha$ by angle in the alternate segment of circle $A K L$ , and so $\angle M D K=\angle L D M=\frac{\pi}{2}-\alpha$ , which thanks to cyclic quadrilaterals gives $\angle M B K=\angle L C M=\frac{\pi}{2}-\alpha$ . From this, we have $B M \perp A C$ and $C M \perp A B$ , and so $M$ indeed is the orthocentre of $A B C$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.