## Thursday, December 2, 2010

### Problem 546: Triangle, Angle Trisectors, 60 Degrees, Equilateral triangle

Geometry Problem
Click the figure below to see the complete problem 546 about Triangle, Angle Trisectors, 60 Degrees, Equilateral triangle.

1. 1)in triangle AEC
<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
EDsinx=AEsina
4)in triangle CEF
<CEF=x
ECsinc=EFsinx
5)DE=EF=ACsina.sinc/(siny.sinx)
<DEF=60,DEF is equilateral
.-.

2. DEF = 60°, ADE = CFE = 120° - a - c
AD 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°

3. Let CE meet AB at M and let AD and CF meet at N

Now < 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