## Friday, May 13, 2011

### Problem 601: Cyclic Quadrilateral, Incenter, Excenter, Rectangle

Geometry Problem
Click the figure below to see the complete problem 601.

1. http://img27.imageshack.us/img27/6927/probelm601.png

A,E,G are collinear and cut arc BC at the midpoint of arc BC.
D,F,H are collinear and cut arc BC at the midpoint of arc BC.
Let L, N, K are the projection of E, M, G over BC . N is the midpoint of BC.
We have BL= p-AC and CK= p-AC where p is half of perimeter of triangle ABC
So N is the midpoint of LK and M is the midpoint of EG.
Similarly M is also the midpoint of HF.
Triangles EBG, ECG, HBF, FCH are right triangles so we have MB=MC=ME=MG=MH=MF
Quadrilateral HGFE is concyclically with diameters HF and EG so EFGH is a rectangle.

Peter Tran

2. Please elaborate how "M is the midpoint of EG"

3. http://img685.imageshack.us/img685/673/probelm6011.png

Draw EQ//BC and MN, GK intersect EQ at P and Q.
Since N is the midpoint of LK so P and M are the midpoint of EQ and EG.

Peter Tran

4. can u say how CK=P-AC