tag:blogger.com,1999:blog-6933544261975483399.post8362620412211977711..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1136: Triangle, Circumcircle, Orthocenter, Midpoint, Arc, 90 Degrees, AngleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-13120384245261167402018-10-26T04:18:35.206-07:002018-10-26T04:18:35.206-07:00Extend HM to X such that HM = MX, so AHCX is a par...Extend HM to X such that HM = MX, so AHCX is a parallelogram<br /> <br />Hence < AHC = A + C = < AXC, hence X must lie on circle ABC {B + (A+C) = 180}<br /> <br />Therefore BX is a a diameter passing thro O and so < BDH = 90<br /> <br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17697078210017377532015-07-15T14:42:26.447-07:002015-07-15T14:42:26.447-07:00http://s17.postimg.org/3wo281qsv/pro_1136.png
Dra...http://s17.postimg.org/3wo281qsv/pro_1136.png<br /><br />Draw diameter BE ; connect AE and EC<br />EC⊥BC => EC//AH<br />EA⊥BA => EA//CH<br />So AHCE is a parallelogram => H, M, E are collinear<br />So BD⊥ MD<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com