|
Let AD=CD=x. Draw a line perpendictular to CD which passes through A. Let P be the point where the perpendictular meets CD. The length of DP is (x-1)/2.
CMN is an isosceles triangle. Draw the altitude perpendictular to MN and let Q be where the altitude meets MN. The length of CM is x/2 and the length of MQ is 1/2.
Let angle ADP (same as ADC) equal t. Angle QCM then is t/2. From triangle ADP, cos(t) = ((x-1)/2)/x = (x-1)/(2x). From triangle QCM, sin(t/2) = (1/2)/(x/2) = 1/x.
Rewriting cos(t) in terms of 1/x: cos(t) = (1 - 1/x)/2
Substituting: cos(t) = (1 - sin(t/2))/2
After using trig identities, an quadratic in sin(t/2):
4*(sin(t/2)) - sin(t/2) - 1 = 0
Solving the quadratic: sin(t/2) = (1 +/- sqrt[17])/8
The angle must be positive then sin(t/2) = (1 + sqrt[17])/8
Since sin(t/2) = 1/x, then CD = x = (sqrt[17] - 1)/2 = 1.56155 |
