tag:blogger.com,1999:blog-6933544261975483399.post2327066608027334320..comments2024-09-10T22:02:29.582-07:00Comments on GoGeometry.com (Problem Solutions): Geometry Problem 1130: Cyclic Quadrilateral, Diagonals, Angle Bisector, Congruent AnglesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-61732210562974019822015-12-20T14:26:59.261-08:002015-12-20T14:26:59.261-08:00Let circumcircles of ABE and DCE meet at G'. R...Let circumcircles of ABE and DCE meet at G'. Radical axis of these circles is G'E. Because FB*FA=FC*FD, F is on line G'E. Note that <BAE=<BG'E and <CDE=<CG'E, but since <BAE=<CDE, <BG'E=<CG'E. G' lies on EF so G and G' concur. With this fact, <AGB=<AEB=<CED=<CGD.Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78289586838311973202015-09-01T07:21:48.559-07:002015-09-01T07:21:48.559-07:00This comment has been removed by the author.Amoreirahttps://www.blogger.com/profile/10656936792574375693noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7648452331712468212015-07-11T21:40:38.759-07:002015-07-11T21:40:38.759-07:00To Antonio
Please see my response below.
1. L and...To Antonio<br />Please see my response below.<br /><br />1. L and N are harmonic conjugate with respect to B and C<br />Let circle diameter LN (Apollonius circle of B and C) cut EF at G’.<br />Since G’ is located on Apollonius circle of B and C so G’EF is an angle bisector of angle BG’C and G’EF ⊥ NG’<br />2. N and M are harmonic conjugate with respect to A and D<br />Since NG’⊥ G’M => G’ locate on Apollonius circle of A and D ( diameter MN)<br />And G’M is an angle bisector of ∠AG’D =>angle ∠AG’B congruent to ∠CG’D.<br /><br />G’ is an only point of line EF have this property since G’ is the perpendicular projection of N over EF. G’ in fact coincide to G .<br /><br />Conclusion: for a general quadrilateral ABCD, we will have one and only one point G on the EF so that EF is the angle bisector of angles BGC and AGD. At the point G , we will have angle AGB congruent to CGD.<br />See link below for the typical non-cyclic quadrilateral ABCD and location of point G satisfied the problem statement<br /><br />http://s12.postimg.org/9cy9rp26l/pro_1130test.png<br /><br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-558447695011693162015-07-10T14:50:52.319-07:002015-07-10T14:50:52.319-07:00Problem 1130 to Peter: in your 2nd statement, why ...Problem 1130 to Peter: in your 2nd statement, why NG is perpendicular to EF? Thanks.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57342575125002565192015-07-09T17:37:24.950-07:002015-07-09T17:37:24.950-07:00Per my solution as above , qudrilateral ABCD may ...Per my solution as above , qudrilateral ABCD may not need to be cyclic to get the result<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77215893351286399492015-07-09T11:36:56.848-07:002015-07-09T11:36:56.848-07:00http://s4.postimg.org/9k94xkj65/pro_1130.png
Let ...http://s4.postimg.org/9k94xkj65/pro_1130.png<br /><br />Let AD cut BC at N and EF cut BC and AD at L and M<br />From N make line NG perpendicular to EF<br />Observe that M, N are harmonic conjugate points of A and D and<br />L, N are harmonic conjugate points of B, C<br />So circumcircles of triangles MGN and LGN are Apollonius circles of AD and BC<br />And GM and GL are internal bisectors of angles AGD and BGC <br />So ∠ (AGB)= ∠ (DGC)<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com