Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Wednesday, July 8, 2015

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

Labels:
angle bisector,
congruence,
cyclic quadrilateral,
diagonal

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

ReplyDeleteLet 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)

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

ReplyDeletePeter Tran

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

DeleteTo Antonio

DeletePlease 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

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ReplyDeleteLet 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.

ReplyDelete