tag:blogger.com,1999:blog-6933544261975483399.post5702051201249421400..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 489: Parallelogram, Triangle, Quadrilateral, AreaAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-11040913179579347902010-08-02T00:38:08.764-07:002010-08-02T00:38:08.764-07:00▲AFB ~ ▲BCE => FD/AF = h/h2 (h, h2 alt FDE,...▲AFB ~ ▲BCE => FD/AF = h/h2 (h, h2 alt FDE, FDC )<br />=> FD∙h2 = AF∙h => SFDC = SAFE =><br />SAGCD = SFGEc .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36268522909458256822010-08-02T00:33:17.728-07:002010-08-02T00:33:17.728-07:00Tr. EFC/Tr.EFD = EC/ED = EB/EF as BC//FD. But EB/E...Tr. EFC/Tr.EFD = EC/ED = EB/EF as BC//FD. But EB/EF=DA/DF as AB//DC. So Tr. EFC/Tr.EFD =DA/DF=Tr.EAD/Tr.EFD or Tr. EFC = Tr.EAD and thus Tr. EFC - Tr. EGC = Tr.EAD - Tr.EGC or A2 = A1 as reqd.<br />Vihaan, DubaiVihaanhttps://www.blogger.com/profile/14122321143712979648noreply@blogger.com