Let ABCD be an isosceles trapezoid with AB parallel to CD, angle ADC equal to angle BCD and AD = DC = CB. Let M be the midpoint of BC and N be the midpoint of CD.

If AB=MN=1 then how long is CD?

Let x = |BM| = |CM| = |CN|, y = |BN|, 
    p = angle CNM, and t = angle BNM.

From triangle CNM:

    2*x = 1/cos(p)                                    (1)

From triangle ABN:

    2*y = 1/cos(p+t)                                  (2)

From triangle BNM:

    x/sin(t) = 1/sin(p-t)                             (3)

    y/sin(p) = 1/sin(p-t)                             (4)

Combining (1) and (3) we get

    sin(p-t) = 2*cos(p)*sin(t)

             or

    tan(t) = tan(p)/3                                 (5)

Combining (2) and (4) we get 

    sin(p-t) = 2*cos(p+t)*sin(p)     

             or

    tan(p) - tan(t) = 2*sin(p)*(1 - tan(p)*tan(t))    (6)

Combining (5) and (6) we get

    4*cos(p)^2 - cos(p) - 1 = 0

             or

              1 + sqrt(17)
    cos(p) = --------------
                   8

Therefore,

                        8
    |CD| = 2*x = -------------- ~= 1.56155
                  1 + sqrt(17)

Posted by Bractals on 2006-01-02 12:25:56