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Math Help - Triangles

  1. #1
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    Triangles

    In a triangle ABC, D and F are points on BC, and E is a point on AB such that AD is parallel to EF, AngleCEF = AngleACB, AD = 15m, EF = 8m and BF = 8m. Find the length of BC.
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  2. #2
    MHF Contributor
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    Here is one way.

    Let angle CEF and angle ACB be theta each.

    -------------
    In triangle ADC:

    Let x = side DC.
    By Law of Sines,
    (AD = 15)/sin(theta) = x/sin(angle CAD)
    Cross multiply,
    x*sin(theta) = 15*sin(angle CAD)
    sin(theta) = [15*sin(angle CAD)]/x ----(i)

    -------------
    In triangle EFC:

    FD = 15-8 = 7 -----why?

    EF = BF = 8 ----that means triangle BFE is isosceles.

    AB is parallel to EB; BD is parallel to BF; and AD is parallel to EF. Therefore, triangle BDA is similar to triangle BFE.
    Hence, triangle BDA is also isosceles, and since AD=15, then BD=15 also.
    Since BF=8, then FD = BD-BF = 15-8 = 7 ----***

    So, by Law of Sines again,
    (FC)/sin(theta) = (EF)/sin(angle FCE)
    Substitutions,
    (7+x)/sin(theta) = 8/sin(angle FCE)
    Cross multiply,
    8*sin(theta) = (7+x)*sin(angle FCE)
    sin(theta) = [(7+x)sin(angle FCE)]/8 ----(ii)

    ------------
    sin(theta) from (i) = sin(theta) from (ii),
    [15sin(angle CAD)]/x = [(7+x)sin(angle FCE)]/8
    Cross multiply,
    x[(7+x)sin(angle FCE)] = 8[15sin(angle CAD)]
    (7x +x^2)sin(angle FCE) = 120sin(angle CAD) ----(iii)

    You have noticed maybe that I am about to show that (angle FCE) = (angle CAD).

    In the figure, in triangles EFC and ADC,
    Since EF and AD are parallel, then (angle EFC) = (angle ADC).
    Given: (angle CEF) = (angle ACB, same as angle ACD)
    That means: (two angles of triangle EFC) = (two angles of triangle ADC).
    That means further that their third angles are equal also.
    Hence, (angle FCE) = (angle CAD) ----***

    Substitute that into (iii),

    (7x +x^2)sin(angle FCE) = 120sin(angle CAD) ----(iii)
    (7x +x^2)sin(angle CAD) = 120sin(angle CAD)
    7x +x^2 = 120
    x^2 +7x -120 = 0 ----(iv)

    Factor that, or use the quadratic formula, and you will get x = 8.
    Therefore,
    BC = BF +FD +x
    BC = 8 +7 +8
    BC = 23 ----answer.
    Last edited by ticbol; June 17th 2005 at 11:24 PM.
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  3. #3
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    Thanks a lot, that is excellent, I could not find a way to prove that triangle AEC was congruent to triangle CEF.

    Although i did get a ratio: where y = CD

    Then (y+7)/15 = y/8

    y = 8m

    Thanks once again!
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  4. #4
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    One more thing ticbol, I love your style in working out and providing very throuogh answers. Also the writing of 'answer' at the end is also a fond habit of mine in all exams.

    Once again keep up the fab work.
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