Question 21
(IRE 1) 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) $\frac{AM}{AX}=\cos\angle BAC$.
(a) $\angle BAM=\angle CAX$, and
(b) $\frac{AM}{AX}=\cos\angle BAC$.