Geometry Problem

Click the figure below to see the complete problem 538 about Triangle, Perpendicular Bisector, Circumcircle, Midpoint.

Go to Complete Problem 538

## Saturday, November 6, 2010

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

Labels:
circle,
circumcircle,
midpoint,
perpendicular,
triangle

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Denote (XYZ) =angle(XYZ)

ReplyDeleteDraw 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

Peter:

ReplyDeleteAfter 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]

César Lozada

To Cesar

ReplyDeleteIn the 1st part of my solution, by drawing circles F and H we have FD=FN' and HD=HN'( N and N' are 2 separated points)

Note that N is the intersecting point of DM to circumcircle of triangle ABC while N' is the intersecting point of circle F to circle H . I try to prove that these 2 points N and N' coincide . If these 2 points coincide then FH become perpendicular bisector of DN . Hope that It will clear any confusion .

Peter Tran

Peter Tran