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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Details: Click on the figure below.
P, P' tg points ( on AC, BC). Join O2 to CFrom AD=AE, PC=P'C, we get PC=11Right tr O1PP'~tr O2CE => 6/11=7/x => x=77/6
https://goo.gl/photos/gUJk9ftBfbk8pmpQ9Define points I, F, G, H and M as per sketchWe have CG=CE=BF= 7 and BG=CF=CH=11We also have IC⊥CMTriangle IHC simillar to CEM ( case AA)So IH/CH= CE/ME => 6/11=7/xSo x= 77/6
If U,V,W are the tangency points of BC, CA, AB respectively of the in circle, Then WD = VE = BC = 18.So easily VC = 11Hence from similar right triangles O1VC and O2CE 6/11 = 7/x and so x = 77/6Sumith PeirisMoratuwaSri Lanka