Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it China

Figure 1
From point PP outside a circle draw two tangents to the circle touching at points AA and BB. Draw a secant line intersecting the circle at points CC and DD, with CC between PP and DD. Choose point QQ on the chord CDCD such that DAQ=PBC\angle DAQ = \angle PBC. Prove that DBQ=PAC\angle DBQ = \angle PAC.

Solution

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

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

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