tag:blogger.com,1999:blog-6933544261975483399.post4968008021093665957..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 521 Triangle, Squares, Altitude, Rectangles, AreasAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-9827655679589725652010-09-13T07:33:27.045-07:002010-09-13T07:33:27.045-07:00ABH = ACP => ▲ABH ~ ▲ACP => AB/AC = AH/AP ...ABH = ACP => ▲ABH ~ ▲ACP => AB/AC = AH/AP =><br />AB∙AP = AC∙AHc .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90054271438087248492010-09-12T23:58:38.515-07:002010-09-12T23:58:38.515-07:00We have Angle AHB=angle AMB =90 and quadrilateral ...We have Angle AHB=angle AMB =90 and quadrilateral ABMH is cyclic<br />CMB and CHA are 2 secants from point C to circle ABMH<br />So CM.CB=CH.CA <br />But CM.CB=S2’ and CH.CA=S2 and S2=S2’<br />Similarly we also have S1=S1’ and S3=S3’<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com