tag:blogger.com,1999:blog-6933544261975483399.post2040668696423100656..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 733: Triangle, Orthocenter, Altitude, Reflection in a line, Circumcircle, ConcurrencyAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-28100322658807302882015-09-12T15:06:30.648-07:002015-09-12T15:06:30.648-07:00http://s15.postimg.org/lsh45cs6j/Problem_733.png
S...http://s15.postimg.org/lsh45cs6j/Problem_733.png<br />See link above for the sketch<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44417554213814323152015-09-12T04:35:30.564-07:002015-09-12T04:35:30.564-07:00i cannot open the attached sketch..plz open this f...i cannot open the attached sketch..plz open this for...i just need this for my presentation.<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-82191241660039038252012-02-27T17:35:18.695-08:002012-02-27T17:35:18.695-08:00http://img718.imageshack.us/img718/9332/problem733...http://img718.imageshack.us/img718/9332/problem733.png<br />below is my comment with some typo correction and clarification.<br />Let BH, AH and CH cut the circle at B’, A’ and C’ and A, B, C are angles of triangle ABC<br />Let angles x, y, z denotes angles shown on attached sketch.<br />A’, B’, C’ are points of symmetric of H over BC, AC and AB ( property of orthocenter) <br />So ∠(C’AB’)=2(∠BAH+∠HAC)= 2∠ A<br />1. Let P is the intersection of C’C1 to B’B1<br />In triangle C1PB1 ∠ ( C1PB1)= 2(y+z)-180= 180-2∠A<br />So quadrilateral AC’PB’ is cyclic => point P will be on the circle.<br /><br />2. Let P1 is the intersection of A1A’ to C1C’<br />∠ (C’BA’)= 2∠ (B) <br />In triangle BA1C1 external angle B=x+z<br />In Triangle A1C1P1, angle(C1P1A1)= 180-2x-2z=180-2B<br />So quadrilateral C’BA’P1 is cyclic => point P1 will be on the circle<br />Both points P and P1 are the intersection of line C2 to the circle => P coincide to P1<br /><br />So A1A2, B1B2 , C1C2 are concurrent at P on circlePeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com