+819 In triangle $ABC$, let the perpendicular bisector of $BC$ intersect $BC$ and $AC$ at $D$ and $E$, respectively. If $BC = 20$ and $\angle C = 15^\circ$, then find the length of $BE$.
+128656 
Triangle ABC is isosceles with AB = AC
Angle ADB = 90
Angle ABD = 15
Angle BAD = 75
AB = 10 / sin 75° = 10 / cos 15°
AE = EC = 5 / cos 15°
BE^2 = BC^2 + EC^2 - 2 ( BC * EC) cos (15°)
BE^2 = 20^2 + 25 / [cos 15°]^2 - 2 ( 20 * 5/ cos (15°) * cos 15°
BE^2 = 400 + 25 / [ cos 15°]^2 - 200
BE^2 = 200 + 25 / [ cos 15°]^2 { [ cos 15°]^2 = (2 + sqrt 3)/4
BE^2 = 200 + 25 * 4 / ( 2 + sqrt 3)
BE^2 = 200 + 100 / (2 + sqrt 3)
BE^2 = 100 ( 2 + 1/ (2 + sqrt 3) )
BE^2 = 100 ( 2 + 2 - sqrt 3 )
BE^2 = 100 ( 4 -sqrt 3)
BE = 10 sqrt ( 4 - sqrt 3)
