Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

In triangle ABC, the bisectors of <B and <C meet AC and AB at M and N respectively.?

If segments BM and CN have the same length, prove that triangle ABC is isosceles.

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  • Anonymous
    1 decade ago
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    I call : AB = c, BC = a , CA = b

    AN = dA, BM = dB,

    AM=b1, MC=b2

    BN=a1, NC=a2

    Now two theorems of plane geometry say:

    1) d²B = ac - b1*b2

    d²A = bc - a1.a2

    2) c : b = a1 : a2 or a1.c=a2.b

    a : c = b2 : b1 or b2.c = b1.a

    Given is:

    d²(A) = b.c - a1.a2 = d²(B) =c.a - b1.b2

    (b1+b2).c - a1.a2 = c(a1+a2) - b1.b2

    b1.c + b2.c -a1.a2 = c.a1 + c.a2 - b1.b2

    Now with b2.c = a.b1 and c.a2 = a1.b we have:

    b1.c + a.b1 -a1.a2 = c.a1 + a1.b -b1.b2

    c.b1 + a1.b1 + a2.b1 - a1.a2 = c.a1 + a1.b1 + a1.b2 - b1b2

    c.b1 + a1.b1 + a2.b1 + b1.b2 = c.a1 + a1.b1 + a1.b2 + a1.a2

    c.b1 +a2.b1 + b1.b2 = c.a1 +a1.b2 + a1.a2

    b1(c+a2+b2) = a1(c+b2+a2)

    So a1=b1 and ABM and ABN are congruent : <A = <B

    Previewing what I did, I saw that I should have the lettering of the triangle wrong, sorry for that. (HvD)

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