tag:blogger.com,1999:blog-6933544261975483399.post1947844898883843036..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 481: Triangle, Altitude, Cevian, Perpendicular, AngleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-28321424427498000362017-04-02T09:52:44.156-07:002017-04-02T09:52:44.156-07:00ABFE is cyclic ==> <AFE = <FBG and we c...ABFE is cyclic ==> <AFE = <FBG and we conclude that <ABE = <CBG.Ignacio Larrosa Cañestrohttps://www.blogger.com/profile/03014700623279846626noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48898763518754445502016-06-21T02:48:22.969-07:002016-06-21T02:48:22.969-07:00Let BG and AF meet at X
Then < ABE = < AFE ...Let BG and AF meet at X<br /><br />Then < ABE = < AFE = < XFG = < FBX<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-43268812883354892392010-07-27T20:32:37.701-07:002010-07-27T20:32:37.701-07:00I think its still easier...
AEFB cyclic implies t...I think its still easier...<br /><br />AEFB cyclic implies that <ABE = <AFE=90-<BFG=90-(90-<FBG)=<FBG<br /><br />as dessired.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-20211915481610855842010-07-27T10:30:50.859-07:002010-07-27T10:30:50.859-07:00Let (XYZ) denote angle XYZ
Note that quadrilateral...Let (XYZ) denote angle XYZ<br />Note that quadrilateral ABFE is cyclic . (AEB)=(AFB)=90<br />We have (BAF)=(BEF) ( both angle face the same arc BF)<br />Angle alpha = (ABF)- (EBF) and angle beta= (EBG)-(EBF) <br />But ( ABF)= (EBG) ( both angle complementary to (BAF) and ( BEF) )<br />So alpha=beta<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com