A certain triangle ABC has D on AB and E on AC such that AD=DB=BC=CE and AE=ED. What is the measure of angle BAC?

Let x = AD = DB = BC = CE, y = AE = ED, and
    z = measure of angle BAC.

Applying the cosine rule to triangle DAE,

    y^2 = x^2 + y^2 - 2xy cos(z)

                 or

    x = 2y cos(z)                                   (1)                   

Applying the cosine rule to triangle BAC,

    x^2 = (2x)^2 + (x+y)^2 - 2(2x)(x+y) cos(z)

                 or

    0 = 4x^2 + 2xy + y^2 - 4x(x+y) cos(z)           (2)

Substituting x from (1) into (2) we get

    16 cos(z)^3 - 8 cos(z)^2 - 4 cos(z) - 1 = 0

Solving for z using MathCad 12 we get

    z ~= 29.5467128538 degrees

Posted by Bractals on 2005-10-24 20:48:48