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Prove that BAM=CAX and AM/AX=cos(BAC)

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August 29, 2010
geometrycircumcircletrigonometryTriangleIMO Shortlist

Problem Statement

The tangents at BB and CC to the circumcircle of the acute-angled triangle ABCABC meet at XX. Let MM be the midpoint of BCBC. Prove that
(a) ∠BAM=∠CAX\angle BAM = \angle CAX, and
(b) AMAX=cos⁡∠BAC.\frac{AM}{AX} = \cos\angle BAC.
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