## Wednesday, September 28, 2016

### Geometry Problem 1266 Triangle, Excircle, Circle, Tangency Points, Perpendicular, 90 Degrees, Angle Bisector

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1266.

#### 3 comments:

1. https://goo.gl/photos/ywDg9feFQ4TrtjkP6

Let ED meet AC at P.
Apply Menelaus’s theorem in secant EDP of triangle ABC
PC/PA x EA/EB x DB/DC= 1
Replace EB=DB in above expression and simplify it . we get PC/PA=DC/EA= FC/FA
So (ACFP)= -1
Apollonius circle with diameter FP will pass through G
So GF is an angle bisector of angle AGC

2. If 2s = a+b+c with the usual notation,
AE = AF = s-c and CF = CD = s-c.

If FD = m and FE = n and if X is the midpoint of FD then Tr.s GXD and EAF are similar being isoceles and having equal angles A/2

Hence (m/2)/GD = (s-c)/n ....(1)
Similarly (n/2)GE = (s-a)/m....(2)

From (1) and (2) we have (s-a)/(s-c) = GD/GE and since < BED = < BDE ( each angle = 90-B/2), Tr.s AEG and CDG are similar.

So < AGE = < CGD, hence GF bisects < AGC

Sumith Peiris
Moratuwa
Sri Lanka

3. Problem 1266
Is triangleFEG similar triangleCOD(<CDO=90=<FGE,<FEG=<FOD/2=<COD) so FG/CD=EG/OD.
But triangleFGD is similar with triangle AEO then FG/AE=GD/EO. By dividing by members we have AE/CD=EG/GD(OE=OD). Is BE=BD so <AEG=<CDG, therefore triangleAEG is similar with
triangle CDG.So<AGE=<CGD.Then <FGA=<FGC (as complementary angles ).
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE