Proposed Problem

Click the figure below to see the complete problem 366 about Scalene triangle, Circumcircle, Angles, 60 Degrees, Equilateral triangle.

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Complete Geometry Problem 366

Level: High School, SAT Prep, College geometry

## Thursday, October 15, 2009

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

Labels:
60,
angle,
circumcircle,
equilateral,
scalene,
triangle

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

ReplyDeleteE1 = 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

Let < ADF = < ACF =x, <ADE = <ABE = y, <BEF = BCF = z, <CBE = <CFE = u,

ReplyDelete<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

https://photos.app.goo.gl/7pBHWGa7akiE41W46

ReplyDeleteconnect BD , BF

we have ^(BPA)=60 +^(C)= ^(BDA)+^(DBE)=^(C)+^(DBE)

so Arc(DE)= 120 degrees

similarly we also have Arc(EF)= 120 degrees

so triangle DEF is equilateral