tag:blogger.com,1999:blog-6933544261975483399.post6873043757033150423..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1280 Quadrilateral, Perpendicular Diagonals, Isosceles Right Triangles, 45 Degrees, Collinear PointsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-81876828324214959802016-10-30T01:18:46.078-07:002016-10-30T01:18:46.078-07:00∵ The diagonals AC and BD are perpendicular at E
∴...∵ The diagonals AC and BD are perpendicular at E<br />∴ ∠BEC = 90°<br />∵ AFD and BGC are isosceles right triangle<br />∴ ∠BGC = 90°<br />∴ ∠BEC = ∠BGC<br />∴ B, E, G and C are concyclic<br />∴ ∠GEC = ∠GBC<br />∵ AFD and BGC are isosceles right triangle<br />∴ ∠GBC = ∠GCB<br />∠GBC + ∠GCB + ∠BGC = 180°<br />2∠GBC + 90° = 180°<br />∠GBC = 45°<br />∴ ∠GEC = 45°<br />Similarly, ∠DEF = 45°<br />∵ The diagonals AC and BD are perpendicular at E<br />∴ ∠CED = 90°<br />∠DEF +∠CEF = 90°<br />45° +∠CEF = 90°<br />∠CEF = 45°<br />∵ ∠CEG = ∠CEF<br />∴ The points E, F, and G are collinearAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88261021279188545802016-10-29T22:49:17.933-07:002016-10-29T22:49:17.933-07:00Suppose that F is not on EG and that F' is the...Suppose that F is not on EG and that F' is the point at which AD subtends 90 on line EG. EF'G are collinear. <br />Let X be any point on FE extended.<br /><br />Since BCGE is concyclic < XEB = 45<br />Hence < XEA = 45 = < F'DA since AEF'D is concyclic.<br /><br />So F and F' must coincide and E,F,G are hence collinear <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70330790713314583182016-10-29T21:59:36.083-07:002016-10-29T21:59:36.083-07:00Problem 1280
Is <AED=AFD=90 so AEFD is cyclic,...Problem 1280<br />Is <AED=AFD=90 so AEFD is cyclic,then <CEF=<FDA=45. But <BEC=<BGC=90 so BEGC is cyclic, then <GEC=<GBC=45=<CEF .Therefore the points E,F and G are collinear.<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88770458333153079792016-10-29T20:34:25.066-07:002016-10-29T20:34:25.066-07:00https://goo.gl/photos/3q9h6djv9nLppAob9
Let M and...https://goo.gl/photos/3q9h6djv9nLppAob9<br /><br />Let M and N are the midpoint of BC and AD<br />We have MG=MB=MC=ME<br />And NF=NA=ND=NE<br />So M and N are the centers of quadrilateral BEGC and AEFD<br />In qua. BEGC we have ∠CEG=∠CBG= 45 degrees <br />In qua. AEFD we have ∠FED=∠EAD= 45 degrees<br />So F and G are located on angle bisector of angle CED => E, F and G are colinnear<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com