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.
Saturday, April 5, 2014
Geometry Problem 1001: Triangle, Circumcircle, Perpendicular, Perpendicular Bisector, Tangent, Collinear Points
Subscribe to:
Post Comments (Atom)
ReplyDeleteLet O is the center of the circle
Let the tangent at B cut P1P2 at P
We 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
ReplyDeleteBD is the altitude from B of triangle ABC . D is on AC
DeleteTriangle (BAE) ... What is E?
ReplyDeleteBe is a diameter of circumcircle of triangle ABC
Delete