+153 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 $BE$.
+129771 A E 75°
B D C15°
10 10
Since ED is a perpendicular bisector... BD / CD = BE / CE
BD = CD so BE = CE
CE = DC / sin DEC
CE = 10 / sin (75°)
CE = 10 / sin (30 + 45)
CE =10 / [ sin30 cos 45 + sin45 cos 30 ]
CE = 10 [ (1/2)(sqrt 2/2 + sqrt2/2sqrt 3/2]
CE =10 [ sqrt 2/4 + sqrt 6/4 ]
CE = 40 / [ sqrt 2 +sqrt 6]
CE = 40 [ sqrt 6 -sqrt 2 ] / [ 6 -2]
CE = 10 [ sqrt 6 -sqrt 2] ≈ 10.35 = BE
