## Wednesday, July 8, 2015

### Geometry Problem 1130: Cyclic Quadrilateral, Diagonals, Angle Bisector, Congruent Angles

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.

1. http://s4.postimg.org/9k94xkj65/pro_1130.png

Let AD cut BC at N and EF cut BC and AD at L and M
From N make line NG perpendicular to EF
Observe that M, N are harmonic conjugate points of A and D and
L, N are harmonic conjugate points of B, C
So circumcircles of triangles MGN and LGN are Apollonius circles of AD and BC
And GM and GL are internal bisectors of angles AGD and BGC
So ∠ (AGB)= ∠ (DGC)

2. Per my solution as above , qudrilateral ABCD may not need to be cyclic to get the result

Peter Tran

1. Problem 1130 to Peter: in your 2nd statement, why NG is perpendicular to EF? Thanks.

2. To Antonio
Please see my response below.

1. L and N are harmonic conjugate with respect to B and C
Let 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 D
Since 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 statement

http://s12.postimg.org/9cy9rp26l/pro_1130test.png

3. This comment has been removed by the author.

4. 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.