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China Mathematical Competition (Extra Test)

Geometry Difficulty 4.9 AIME Prove it China

From point $P$ outside a circle draw two tangents to the circle touching at points $A$ and $B$ . Draw a secant line intersecting the circle at points $C$ and $D$ , with $C$ between $P$ and $D$ . Choose point $Q$ on the chord $CD$ such that $\angle DAQ = \angle PBC$ . Prove that $\angle DBQ = \angle PAC$ .

Solution

Using $\angle DAB = \angle DCB$ , $\angle DAB = \angle DAQ + \angle QAB$ , $\angle DCB = \angle PBC + \angle BPQ$ , and $\angle DAQ = \angle PBC$ , we get

$\angle QAB = \angle BPQ$ , so points $P, A, Q, B$ share a common circle. Then $\angle BQP = \angle PAB$ , that is, $\angle DBQ + \angle CDB = \angle PAC + \angle CAB$ . Since $\angle CDB = \angle CAB$ , so $\angle DBQ = \angle PAC$ .

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