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 1 of 6: Choose the arc point N

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\angle BIC=90^\circ+\frac12\angle A=120^\circ=\angle BNC

Detailed analysis

Let $D=AH\cap BC$ , let $K$ be the second intersection of $AH$ with the circumcircle, and let the line through $I$ perpendicular to $BC$ meet $BC$ at $E$ and the minor arc $BC$ at $N$ . Since $\angle A=60^\circ$ , $\angle BIC=90^\circ+\frac12\angle A=120^\circ$ , while the inscribed angle gives $\angle BNC=180^\circ-\angle A=120^\circ$ .