Problem

• AMB is an isosceles triangle and therefore MA = MB. Also the sizes of angles MAD and MBC are equal. Hence triangles DAM and CBM have an equal angle between two equal sides are therefore congruent. Since triangles DAM and CBM are congruent sides DM and CM are equal in size and therefore triangle DCM is isosceles.
  
  
• We draw the perpendicular MM' to CB and we consider the right triangles MCM' and MBM' and the tangent of angles MCM' and MBM'.
  tan(MCM') = MM' / CM'
  tan(MBM') = MM' / BM'
  
• The above may be written as
  CM' = MM' / tan(MCM') and BM' = MM' / tan(MBM')
  
• Note that the size of angle MBM' is equal to 75 degrees. Let the length of the side of the square be x. Hence
  MM' = x / 2 and CM' + BM' = x
  
• Substitute the above in the equations CM' = MM' / tan(MCM') and BM' = MM' / tan(MBM') to obtain
  x = (x/2) / tan(MCM') + (x/2) / tan(75 degrees)
  
• Divide all terms of the equation by x and multiply them by 2 to obtain
  2 = 1 / tan(MCM') + 1 / tan(75 degrees)
  
• tan(75 degrees) is calculated using the tangent of a sum formula follows
  tan(75 degrees) = tan(30 degrees + 45 degrees) =
  [ tan(30) + tan(45) ] / [ 1 - tan(30)*(tan(45) ] = 2 + sqrt(3)
  
• Substitute tan(75 degrees) by 2 + sqrt(3) into the equation 2 = 1 / tan(MCM') + 1 / tan(75 degrees) to obtain
  1 / tan(MCM') = 2 - 1 / [ 2 + sqrt(3) ]
  
• Simplify the above and solve for tan(MCM') to obtain
  tan(MCM') = 1 / sqrt(3)
  
• Solve the above trigonometric eqution to find the size of angle MCM'
  MCM' = 30 degrees
  
• The size of angle DCM is now calculated as follows
  DCM= 90 degrees - 30 degrees = 60 degrees
  
• and since triangle DCM is isosceles then angle CDM is also equal to 60 degrees and the triangle is equilateral.

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