See complete Problem 208 at:
gogeometry.com/problem/p208_triangle_excircles_angles.htm
Triangle, Excircles, Angles, 360 degrees. Level: High School, SAT Prep, College geometry.
Sunday, November 16, 2008
Elearn Geometry Problem 208: Triangle, Excircles, Angles, 360 degrees
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with the exteror angles of ABC ,and isoscele triangles like BEF we have:
ReplyDeletealpha = 180-(B/2+C/2)
beta = 180-(C/2+A/2)
gamma = 180-(A/2+B/2)
sum = 3.180-180=360
.-.
Let :
ReplyDelete∠ABC=s
∠ACB=u
∠BAC=t
Let O be the center of the circle E_a. Then join DO and OF.
Extend DO to meet the circle at the point P
∠DPF=180-α
therefore
∠DOF = 360-2α
DOFC lies on a circle therefore
(1.) 180+u=2α
Working in the same way we can find similar formules for β and θ
(2.)180+t=2β
(3.)180+s=2θ
Adding equations 1 ,2 and 3 with u+s+t=180
2(α+β+θ)=540+s+t+u
2(α+β+θ)=720
α+β+θ=360
since F,E and G,H are tangency points therfore : BF=BE , BG=BH
ReplyDeleteand similarly we have : CL=CK , CH=CJ and AE=AD , AM=AL
then we conclude the following results:
(DE)//(ML) , (FE)//(GB) ,(LK)//(HJ)
hence if we matching line (DE) with (ML)
line (HG) with (EF)
line (BJ) with (LK)
we have: α+β+θ=360
Consider the angles at E, H, L respectively
ReplyDeleteA/2 + B/2 + Alpha = 180
B/2 + C/2 + Beta = 180
C/2 + A/2 + Theta = 180
Adding A + B + C + Alpha + Beta + Theta = 540, from which
Alpha + Beta + Theta = 540 - 180 = 360
Sumith Peiris
Moratuwa
Sri Lanka