tag:blogger.com,1999:blog-6933544261975483399.post2216778350364014755..comments2024-10-11T10:08:49.545-07:00Comments on GoGeometry.com (Problem Solutions): Geometry Problem 1459: Two Triangles, Orthocenter, Midpoint, PerpendicularAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-71754109433175171772020-03-11T08:22:51.983-07:002020-03-11T08:22:51.983-07:00Let circumradius of Tr.AED be R
Then AG=2R.cos(A)=...Let circumradius of Tr.AED be R<br />Then AG=2R.cos(A)=(ED/sin(A)).cos(A)=ED/tan(A)<br />Similarly in Tr. ABC AF= BC/tan(A),<br />We have AG/AF = ED/BC. <br />Let midpoint of BD is J. Then MJ=ED/2 and NJ=BC/2. <br />MJ/NJ=ED/BC, also MJ and NJ are parallel to ED and BC respectively. <br />We get MJ/NJ=AG/AF, AG and AF are 90 degree clock wise rotation of MJ and NJ.<br />Hence Tr.MJN and Tr.GAF are similar, and GF is 90 degree clockwise rotation of MN. <br />Hence MN is perpendicular to FG. Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.com