Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
http://s4.postimg.org/9k94xkj65/pro_1130.pngLet AD cut BC at N and EF cut BC and AD at L and MFrom N make line NG perpendicular to EFObserve that M, N are harmonic conjugate points of A and D andL, N are harmonic conjugate points of B, CSo circumcircles of triangles MGN and LGN are Apollonius circles of AD and BCAnd GM and GL are internal bisectors of angles AGD and BGC So ∠ (AGB)= ∠ (DGC)
Per my solution as above , qudrilateral ABCD may not need to be cyclic to get the resultPeter Tran
Problem 1130 to Peter: in your 2nd statement, why NG is perpendicular to EF? Thanks.
To AntonioPlease see my response below.1. L and N are harmonic conjugate with respect to B and CLet circle diameter LN (Apollonius circle of B and C) cut EF at G’.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’2. N and M are harmonic conjugate with respect to A and DSince NG’⊥ G’M => G’ locate on Apollonius circle of A and D ( diameter MN)And G’M is an angle bisector of ∠AG’D =>angle ∠AG’B congruent to ∠CG’D.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 .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.See link below for the typical non-cyclic quadrilateral ABCD and location of point G satisfied the problem statementhttp://s12.postimg.org/9cy9rp26l/pro_1130test.png
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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.