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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1184.
Let r1 and r2 are radii of circles Q and OSince TH is the common tangent of 2 circles so we have PH/PT= PQ/PO= r1/r2Let AC meet OQ at P’We have AO ⊥OQ and CQ⊥OQSo P’C/P’A= QC/OA=r1/r2=P’Q/P’OSo P coincide to P and TH, EF and AC concur at P
Ang CFB=45=P1+FCP, Ang BFG=45=P2+FMP(M on GF)Ang FMP=CNP (N on DB) => in tr PNC = 180
OQ, TH meet at P. We must prove that AC also passes thro PIf OT = R and QH = r, then OA = sqrt2 R and QC = sqrt2 rSo OA/QC = OT/QH which in turn = PO/PQ Now consider the right triangles POA and PQC in which PO/PQ = OA/QC and < O < Q = 90It follows that the 2 right triangles are similar and so < PAO =PCQ and for this to happnd PCA must be collinear Hence AC, OQ and TH meet in a point PSumith PeirisMoratuwaSri Lanka