tag:blogger.com,1999:blog-6933544261975483399.post5129943697592227000..comments2024-02-22T05:31:28.964-08:00Comments on Go Geometry (Problem Solutions): Problem 538: Triangle, Perpendicular Bisector, Circumcircle, MidpointAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-9133183710959891792010-11-08T11:59:13.485-08:002010-11-08T11:59:13.485-08:00To Cesar
In the 1st part of my solution, by drawi...To Cesar<br /><br />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)<br />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 .<br />Peter Tran<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69581482005694406422010-11-08T10:38:54.374-08:002010-11-08T10:38:54.374-08:00Peter:
After you draw circles F and H, you have FD...Peter:<br />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]<br /><br />César LozadaAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36553744049806079092010-11-07T13:31:41.263-08:002010-11-07T13:31:41.263-08:00Denote (XYZ) =angle(XYZ)
Draw circle centered F r...Denote (XYZ) =angle(XYZ)<br />Draw circle centered F radius FA=FD and circle centered H radius HD=HC<br />Circle F and circle H intersect each other at D and N’ ( see details picture in the link below)<br />http://img710.imageshack.us/img710/3702/problem538.png<br /><br />N’ is the symmetric point of P over FH . We will prove that N’ coincide to N .<br />1 In triangle ABC , (ABC)=180- (BAC)- (BCA)<br />In circle F , (AN’D)= (AFE) = 90- (BAC)<br />In circle H, (DN’C)= (GHC)=90- (BCA)<br />So (AN’C)= (AN’D) + (DN’C)= 180- (BAC)- (BCA)<br /><br />2 So (ABC)=(AN’C) and quadrilateral AN’BC is cyclic and N’ will be in the circumcircle of triangle ABC<br /><br />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 <br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com