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.

Problem 546: Triangle, Angle Trisectors, 60 Degrees, Equilateral triangle.
Go to Complete Problem 546

Posted by Antonio Gutierrez at 1:40 PM
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3 comments:

  1. AnonymousDecember 3, 2010 at 11:04 AM

    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
    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
    .-.

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  2. c .t . e. oDecember 4, 2010 at 12:12 PM

    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°

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  3. Sumith PeirisOctober 1, 2015 at 8:33 AM

    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

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