Geometry Problem
Click the figure below to see the complete problem 546 about Triangle, Angle Trisectors, 60 Degrees, Equilateral triangle.
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Thursday, December 2, 2010
Problem 546: Triangle, Angle Trisectors, 60 Degrees, Equilateral triangle
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1)in triangle AEC
ReplyDelete<AEC=180-a-c=y
the sine law gives:EC=ACsina/siny;EA=ACsinc/siny
2)at vertex E
<DEF=360-y-60-a-60-c=60
3)in triangle ADE
<ADE=180-a-c=x
EDsinx=AEsina
4)in triangle CEF
<CEF=x
ECsinc=EFsinx
5)DE=EF=ACsina.sinc/(siny.sinx)
<DEF=60,DEF is equilateral
.-.
DEF = 60°, ADE = CFE = 120° - a - c
ReplyDeleteAD meet CF at H => HE bisector
build EAG = c, G on CF, join E to G
▲ADE ≡ ▲EFG (DAE = FGE, AE = EG, AED = GEF )
=> ED = EF => EDF = EFD = 60°
Let CE meet AB at M and let AD and CF meet at N
ReplyDeleteNow < AEM = a+c so < MED = 60 - a, hence < DEF must be 60 (< CEF = 60 + a)
Further < ADE = < EFC = 120 - a - c so < EDN = < EFN
Also E is the incentre of Tr. ANC hence < DNE = < FNE
So Tr. s DEN and FEN are congruent ASA
So in Tr. DEF. DE = EF and the included angle is 60 hence the Tr. DEF is equilateral
Sumith Peiris
Moratuwa
Sri Lanka