## Saturday, August 15, 2009

### Problem 338. Triangle, Circumcircle, Inscribed Circle, Exterior angle bisector, Concyclic points

Proposed Problem
Click the figure below to see the complete problem 338 about Triangle, Circumcircle, Inscribed Circle, Exterior angle bisector, Concyclic points. See more:
Complete Problem 338
Level: High School, SAT Prep, College geometry

1. http://img34.imageshack.us/img34/1809/problem338.png

Connect EA, EB, EDO, DE and EC
Draw a line from O perpendicular to AC. This line intersect circle O at point M.
1. Tri OME and DFE are isosceles with common side ODE and OM//DF
From this we can have m(OEM)=m(DEF) and 3 points E,F and M are collinear .
Since M is the midpoint of arc AC so EM become angle bisector of angle AEC

2. Note that m(AEC)= m(A) +m(C) and m(AEB)= m(C)
So m(BEF)=(m(A)-m(C))/2

3. We have m(BGA)=m(A)- alpha and 2.alpha=m(A)+m(C)
So m(BGA)= (m(A)-m(C))/2 = m(BEF) and quadrilateral GBFE is concyclic

Peter Tran

2. Let the common tangent at E meet AC extended at K.

Let < CKE = 2x so that < FEK = < EFK = 90 - x and let < CEK = y = < CBE.

So < ACE = < ABE = 2x + y and < ABC = 2x + 2y, hence á = 90 - x – y ….(1)

Further < CEF = (90 – x) – y = á = < ABG from (1)

Therefore < EBG = < ABG + < ABE = < CEF + < ACE = < AFE proving that
GBFE is Concyclic

(Note that á = alpha)

Sumith Peiris
Moratuwa
Sri Lanka