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

Labels:
60 degrees,
angle,
equilateral,
triangle,
trisection

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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