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Geometry Problem.Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 1001.
Let O is the center of the circleLet the tangent at B cut P1P2 at PWe have ∆ (BDC) ~ ∆ (BAE)And ∆ (BP1O)~ ∆ (BCA)~ ∆ (BOP2)So BO and BP are angle bisectors of angle P2BP1.Since ∆ (P3CP1) similar to ∆ (P3AP2) => P3P1/P3P2= CP1/AP2 … (1)Since BP is an angle bisector => PP1/PP2= P1B/BP2….. (2)Compare (1) and (2) and note that P1B=P1C and P2A=P2B We have P3P1/P3P2= PP1/PP2 => P coincide to P3 => P1,P2,P3 are collinear
sorry what D
BD is the altitude from B of triangle ABC . D is on AC
Triangle (BAE) ... What is E?
Be is a diameter of circumcircle of triangle ABC