tag:blogger.com,1999:blog-6933544261975483399.post9180176554071822672..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1246: Triangle, Orthocenter, Altitude, Midpoint, PerpendicularAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-60818878776668531862016-08-20T15:09:18.113-07:002016-08-20T15:09:18.113-07:00Thank you Peter for your kind words and for the fi...Thank you Peter for your kind words and for the figure, but I do not know thee<br />the way to move the shape.<br />apostolis manoloudisAPOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60309156611445337092016-08-19T19:05:40.545-07:002016-08-19T19:05:40.545-07:00Excellent solution Apostolis
See link below for th...Excellent solution Apostolis<br />See link below for the sketch of Apostolis's solution<br /><br />https://goo.gl/photos/vsQkRbcfLejMHaTP8<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3464017610077788372016-08-19T01:37:30.232-07:002016-08-19T01:37:30.232-07:00Problem 1246
Let AH1,CH1,EH2 and DH2 intersects ...Problem 1246<br />Let AH1,CH1,EH2 and DH2 intersects the AB,CB at K,L,M and N respectively then <ALC=90=<AKC so ALKC is cyclic. Then AH1.H1K=LH1.H1C. But AH1.H1K is power the point H1 in the circle ( A,E,K)<br />with center M1 and LH1.H1C is power the point H1 in the circle (D,L,C) with center M2.<br />Similar in triangle ADE the point H2 it has an equal forces as to their circles (D,N,C) and (A,E,M) with center M2, M1 respectively .So H1H2 is radical axis of cycles with center Μ1,Μ2.Τherefore H1H2 is perpendicular at M1M2.<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.com