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

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

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## Wednesday, September 28, 2016

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

Labels:
90,
angle bisector,
circle,
excircle,
perpendicular,
tangency point,
triangle

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https://goo.gl/photos/ywDg9feFQ4TrtjkP6

ReplyDeleteLet 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

If 2s = a+b+c with the usual notation,

ReplyDeleteAE = 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

Problem 1266

ReplyDeleteIs 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