Saturday, April 5, 2014

Geometry Problem 1001: Triangle, Circumcircle, Perpendicular, Perpendicular Bisector, Tangent, Collinear Points

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.

Online Geometry Problem 1001: Triangle, Circumcircle, Perpendicular, Perpendicular Bisector, Tangent, Collinear Points

5 comments:


  1. Let 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

    ReplyDelete
  2. Replies
    1. BD is the altitude from B of triangle ABC . D is on AC

      Delete
  3. Triangle (BAE) ... What is E?

    ReplyDelete
    Replies
    1. Be is a diameter of circumcircle of triangle ABC

      Delete