## Thursday, October 15, 2009

### Problem 366. Scalene triangle, Circumcircle, Angles, 60 Degrees, Equilateral triangle.

Proposed Problem
Click the figure below to see the complete problem 366 about Scalene triangle, Circumcircle, Angles, 60 Degrees, Equilateral triangle. See more:
Complete Geometry Problem 366
Level: High School, SAT Prep, College geometry

1. ang C2 = ang E1, ang A1 = ang E2 (same arc FB, BD)

E1 = C2 = 180 - A - 60 - B1 ( from tr BPC )
E2 = A1 = 180 - C - 60 - B2 ( from tr APB )

E1 + E2 = 360 - 120 - ( A + C + B1 + B2 )

E1 + E2 = 60

at the same way F = 60, D = 60 ( ang APC = B + 60 )

E1 = ang FEB, E2 = ang BED

2. Let < ADF = < ACF =x, <ADE = <ABE = y, <BEF = BCF = z, <CBE = <CFE = u,
<CFD = <CAD = v and < BAD = < BED = w

Triangle BPD ; u+z + (A+ 60) = 180

i.e. u+z+v+w = 120….(1) since A = w+z

Similarly w+y+x+z = 120 …(2) since C = z+x

(1) + (2) ; (w+z) + (u+v+w+x+y+z) = 240

But u+v+w+x+y+z = 180

So w+z =60 i.e < E = 60

Similarly < F = 60 and < D = 60

Therefore DEF is equilateral

Sumith Peiris
Moratuwa
Sri Lanka