Geometry Problem

Click the figure below to see the complete problem 601.

## Friday, May 13, 2011

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## Friday, May 13, 2011

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Problem 601: Cyclic Quadrilateral, Incenter, Excenter, Rectangle

## Link List

Online Geometry theorems, problems, solutions, and related topics.

Labels:
cyclic quadrilateral,
excenter,
incenter,
rectangle

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

ReplyDeleteA,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

Please elaborate how "M is the midpoint of EG"

ReplyDeletehttp://img685.imageshack.us/img685/673/probelm6011.png

ReplyDeleteDraw 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

can u say how CK=P-AC

ReplyDelete