Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemClick the figure below to see the complete problem 601.
http://img27.imageshack.us/img27/6927/probelm601.pngA,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 ABCSo 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=MFQuadrilateral HGFE is concyclically with diameters HF and EG so EFGH is a rectangle.Peter Tran
Please elaborate how "M is the midpoint of EG"
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
can u say how CK=P-AC