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.

Problem 538: Triangle, Perpendicular Bisector, Circumcircle, Midpoint
Go to Complete Problem 538


  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)

    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]

    César Lozada

  3. To Cesar

    In 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