tag:blogger.com,1999:blog-6933544261975483399.post1932095308512973147..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 334. Cyclic Quadrilateral, Perpendiculars to DiagonalsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-21395174801738498012011-06-09T10:17:52.094-07:002011-06-09T10:17:52.094-07:00Join MH, GE ...Join MH, GE HMCD is a cyclic quadrilateral with <br />exterior ∠GMH = interior opposite ∠HDC = ∠BDC <br />GFBA is a cyclic quadrilateral with<br />exterior ∠GFH = interior opposite ∠BAG = ∠BAC <br />But ∠BDC = ∠BAC (angles in the same segment) So ∠GMH = ∠GFH and G,F,M,H are concyclicPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-29784758907602023282011-06-08T12:32:59.180-07:002011-06-08T12:32:59.180-07:00To prove that F, M, H, G lie in the same circleTo prove that F, M, H, G lie in the same circleCrisnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59590473133214504972009-12-29T13:18:55.341-08:002009-12-29T13:18:55.341-08:00all tr are similar to each other( ang to perpen si...all tr are similar to each other( ang to perpen sides)<br /><br />AE/AF = ED/DM = EC/CH = BE/BG <br />or BG/AF = CH/DM = ...<br /><br />or BG*DM = AF*CHc .t . e. onoreply@blogger.com