The tangents at $B$ and $C$ to the circumcircle of the acute-angled triangle $ABC$ meet at $X$. Let $M$ be the midpoint of $BC$. Prove that (a) $\angle BAM = \angle CAX$, and (b) $\frac{AM}{AX} = \cos\angle BAC.$
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Tags: geometry, circumcircle, trigonometry, Triangle, IMO Shortlist
The tangents at $B$ and $C$ to the circumcircle of the acute-angled triangle $ABC$ meet at $X$. Let $M$ be the midpoint of $BC$. Prove that (a) $\angle BAM = \angle CAX$, and (b) $\frac{AM}{AX} = \cos\angle BAC.$