## Saturday, November 6, 2010

### Problem 538: Triangle, Perpendicular Bisector, Circumcircle, Midpoint

Geometry Problem
Click the figure below to see the complete problem 538 about Triangle, Perpendicular Bisector, Circumcircle, Midpoint. 1. Denote (XYZ) =angle(XYZ)
Draw circle centered F radius FA=FD and circle centered H radius HD=HC
Circle F and circle H intersect each other at D and N’ ( see details picture in the link below)
http://img710.imageshack.us/img710/3702/problem538.png

N’ is the symmetric point of P over FH . We will prove that N’ coincide to N .
1 In triangle ABC , (ABC)=180- (BAC)- (BCA)
In circle F , (AN’D)= (AFE) = 90- (BAC)
In circle H, (DN’C)= (GHC)=90- (BCA)
So (AN’C)= (AN’D) + (DN’C)= 180- (BAC)- (BCA)

2 So (ABC)=(AN’C) and quadrilateral AN’BC is cyclic and N’ will be in the circumcircle of triangle ABC

3 Both N and N’ are the intersecting points of line DM to circumcircle of ABC so N coincide to N’ and M is the midpoint of DN

Peter Tran

2. Peter:
After you draw circles F and H, you have FD=FN and HD=HN. Then FH is the perpendicular bisector of DN. [A rect if perpendicular bisector of a segment if two points of that rect are equal distanced from the segment extremes]

3. 