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in a triangle ABC it is known that AB=AC. Suppose D is the mid point of AC and BD=BC=2. Then the area of the triangle ABC is

  • math -

    let AD = DC = x
    then AB = 2x
    in triangle ABC , angle B = angle C =Ø
    in triangle BCD , angle C = angle BDC = Ø
    2 angles of one are equal to 2 angles of the other, so they are similar
    then:
    2/x = 2x/2
    2x^2 = 4
    x^2=2
    x=√2
    So triangle ABC has equal sides of 2√2 and a base of 2
    which is similar to
    √2 : √2 : 1
    Using the cosine law we can find angle A
    1^2 = √2^2 + √2^2 - 2(√2)(√2)cosA
    4cosA = 3
    cosA = 3/4
    then sinA = √7/4

    area ABC = (1/2)(2e√2)(2√2)(√7/4)= √7

    check my arithmetic, should have written it out first.

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