Home > Shapes > Geometry
Sum Projection Differences (Posted on 2008-01-05) |
|
Let ABC be an arbitrary triangle.
Let AB and AC be the orthogonal projections of A onto the internal bisectors of angles B and C respectively.
Let BC and BA be the orthogonal projections of B onto the internal bisectors of angles C and A respectively.
Let CA and CB be the orthogonal projections of C onto the internal bisectors of angles A and B respectively.
Prove that |ABAC| + |BCBA| + |CACB| = s,
where s is the semiperimeter of triangle ABC.
Comments: (
You must be logged in to post comments.)
|
Subject |
Author |
Date |
| Nice puzzle! | Chesca Ciprian | 2008-01-07 15:49:37 |
|
|
Please log in:
Forums (0)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
|