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Isosceles Trapezoid (Posted on 2006-01-02) Difficulty: 3 of 5
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?

See The Solution Submitted by Brian Smith    
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Solution Solution | Comment 1 of 3
 

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
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