tag:blogger.com,1999:blog-6933544261975483399.post7057135579084037004..comments2023-03-29T05:17:23.129-07:00Comments on Go Geometry (Problem Solutions): Problem 503: Triangle with Three Squares, Perpendicular, Congruence, ConcurrencyAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-22219788841182065042010-08-16T22:31:23.813-07:002010-08-16T22:31:23.813-07:00There is a minor typo error in my previous comment...There is a minor typo error in my previous comment. Below is the corrected copy<br /><br />Call Q the projection of M over AB<br />We have QM=QB and ang (MQB)=90<br />And QO=QN and ang ( OQN)=90 ( See Problem 497)<br />Triangle MQN congruence with tri. BQO (case SAS)<br />Tri. MQN is the image of Tri. QBO in the rotational transformation of center Q and angle of rotational=90<br />So MN=BO and MN perpen. To BO.<br />With same reason as above we also have MP perpen. To NO and AM perpen. To MO.<br />P is the orthocenter of triangle MNO . AN, BO and CM concurrence at P<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com