tag:blogger.com,1999:blog-6933544261975483399.post5901372095517286852..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 944: Triangle, Orthocenter, Altitude, Perpendicular, 90 Degrees, Parallel LinesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-37981871350448446862013-12-19T14:11:39.294-08:002013-12-19T14:11:39.294-08:00<FHJ=<FEJ=<GKD of cyclic quadrilaterals F...<FHJ=<FEJ=<GKD of cyclic quadrilaterals FJEH and GDKH. But EF//GK so JE//DK.Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79732238575660722232013-12-19T00:32:16.413-08:002013-12-19T00:32:16.413-08:00Correction: FJ//GD in line 2. Correction: FJ//GD in line 2. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-10425528428419051452013-12-18T17:17:36.409-08:002013-12-18T17:17:36.409-08:00Since FE//GK, so HE/HK = HF/HG.
Since FJ//ED, so ...Since FE//GK, so HE/HK = HF/HG. <br />Since FJ//ED, so HF/HG = HJ/HD. <br />Thus, HE/HK = HJ/HD, hence JE//DK. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com