## Saturday, September 24, 2016

### Geometry Problem 1265: Triangle, Incircle, 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 1265.

1. https://goo.gl/photos/mniwJNrj8t6jnXvG7

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. Problem 1265
Is BE=BD then <BED=<BDE or <AEG=<CDG.But <FOC=<FOD/2=<FEG and <CFO=90=<FGE so
Triangle EGF is similar the triangleOFC so EG/OF=GF/FC similar GD/OF=GF/AF with division by states we have EG/GD=AF/FC.So triangleAEG is similar triangleGDC then <EGA=<DGC.
Therefore <AGF=<FGC (as equal angles supplements).
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

3. Really excellent work Apostolis

1. Thank you for your nice words Sumith.

4. Trigonometry solution

EG = 2.AE.sin(A/2).sin(C/2)
GD = 2.CD.sin(C/2).sin(A/2)

So EG/GD = AE/CD
Hence Tr.s AEG and CDG are similar and the result follows

Sumith Peiris
Moratuwa
Sri Lanka

5. Geometry Solution

Let M be the midpoint of EF and let EF =m, FD =n, EG = p and GD = q

Tr.s EGM and DCM are similar (isosceles and both having <C), so
(m/2)/p = (s-c)/n ….(1)
where 2s = a+b+c with the usual notation.

Similarly we can show that
(n/2)/q = (s-a)/m…(2)

(2)/(1) gives, p/q = (s-a)/(s-c)

Hence Tr.s AEG and CDG are similar and the result follows.

Sumith Peiris
Moratuwa
Sri Lanka

6. To prove GF is angle bisector of AngleAGC, it would suffice to prove that the triangles AEG and CDG are similar.
Let A,B,C be the angles of the triangle ABC
Join EF and form the isosceles triangle AEF. Similarly join FD and form the isosceles triangle CDF.

In triangle AEF the angles are A, 90-A/2, 90-A/2
=> Angle GDF = 90-A/2 ( per Ext Segment theorm) => AngleGFD = A/2
Similarly AngleGFE = C/2

Join OE and consider the triangle OAE. The Angles in the triangle are 90,A/2 & 90-A/2 which is similar to the triangle GFD
Hence AE.GD = OE.GF -------------(1)

Similarly join OD and form the triangle ADC whose angles are 90,C/2 & 90-C/2, which is similar to the triangle GFE
Hence EG.DC = OD.GF
=> EG.DC = OE.GF (Since OE=OD) ---------(2)

FROM (1) & (2) AE.GD = EG.DC => Triangles AEG and CDG are similar

Therefore AG/AE = CG/CD
=> AG/AF = CG/CF ----------- (3)

Hence (3) is the result of Angle bisector theorem and GF is the bisector of Angle AGC