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.
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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