Tuesday, May 11, 2010

Problem 450: Triangle, Median, Cevian, Congruence, Proportional Segments

Geometry Problem
Click the figure below to see the complete problem 450 about Triangle, Median, Cevian, Congruence, Proportional Segments.

Problem 450: Triangle, Median, Cevian, Congruence, Proportional Segments.
See also:
Complete Problem 450
Level: High School, SAT Prep, College geometry

5 comments:

  1. draw CM//BD ( M on AE )
    ▲BFE ~ ▲EMC
    EM/y = (a - c)/c => EM = y(a-c)/c
    F midpoint of AM =>
    x = y + EM => x = y + y(a-c)/c => xc = yc + ay - yc
    xc = ay

    x/y = a/c
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  2. by Menelaus' theorem
    AF/FE*EB/BC*CD/AD=1 ⇔
    x/y=a/c

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  3. use sine rule in ▲ABE and ▲FBE to get (sinABD/sinDBC)=(x/y) and then in ▲ABD and ▲CBD to get (sinABD/sinDBC)=(a/c)...equate them to get the result...

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  4. Problem 450
    ABC is the triangle area (ABC).Τhen (ABF)/(BEF)=x/y=(ADF)/(DFE)=(ABD)/(DBE)=υ_1/υ_2.
    (Where υ_1and υ_2 are the heights of the triangles ABD, BED, which correspond to the sides AB, BE).But 1=(ABD)/(BCD)=(c. υ_1)/(a. υ_2) so υ_1/υ_2=a/c.Therefore a/c=x/y.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  5. Draw CG //BD where G is on AE extended

    Tr.s ECG and BFE are similar

    Hence (x-y)/y = (a-c) / c

    So x/y = a/c

    Sumith Peiris
    Moratuwa
    Sri Lanka

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