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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
http://s25.postimg.org/ufbka0myn/pro_1107.pngLet circle D cut AB at E and F Since DC^2=DA.DB => Circle D is Apollonius circle of ABAnd PE and CE are bisectors of angles APB and ACBSo PA/PB=EA/EB=CA/CB (1)Triangles PBC similar to PC1B1 => B1C1= (PC1/PB). BC…..(2)Triangles PAC similar to PC1A1=> C1A1= (PC1/PA). CA …..(3)Divide (2) to (3) and using (1) we get B1C1= C1A1
Let AB cut circle D at Q.Let < BCD = p, <BPD = q, < QPC = r and < QCP = sThen <CAD = p and < PAD = q since DP (=DC) is a tangent to circle APB at P.Further < PDA = 2s and < CDA = 2r so that< PQD = < QPD = 90-s …..(1) and< QCD = CQD = 90-r…….(2)From (2) < ACQ = 90-p-r …..(3)From (1) < APQ = 90-q-s ……(4)Now < ABB1 = 2s+q & …..(5)And < CBA = 2r+p ……(6)Further < PCB = 90-p-r-s from which< A1B1C1 = 90-p-q-r-s …..(7) since < PCB = < PB1C1 and < BB1A = BAA1 = qAlso < B1A1C1 = < PA1C1 - <PA1B1 = < ACC1 - < ABB1 = (90-p-r+s) – (2s+q) =90-p-q-r-s…(8)From (7) and (8), < A1B1C1 = < B1A1C1 and therefore C1A1 = C1B1Sumith PeirisMoratuwaSri Lanka