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

Labels:
circumcircle,
collinear,
perpendicular bisector,
tangent,
triangle

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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