Problem 2

Let

ABC

be an acute triangle with

\angle BAC=60^\circ

and

AB>AC

. Let

I

be its incenter and

H

its orthocenter. Prove that

2\angle AHI=3\angle ABC

Step 3 of 6: Relate H, K, I, and N

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HD=DK,\quad ED\perp IN,\quad ED\perp HK\Longrightarrow IHKN\text{ is an isosceles trapezoid}

Detailed analysis

Because $H$ is the orthocenter, the altitude line $AH$ meets the circumcircle again at $K$ with $HD=DK$ . Since $ED\perp IN$ and $ED\perp HK$ , the quadrilateral $IHKN$ is an isosceles trapezoid, so $IH=NK$ and the corresponding angles are equal.