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 $BE$.
CD = BD = 10
ED = ED
Thus by LL , triangle CDE is congruent to triangle BDE
So CE = BE
sin CED / CD = sin CDE / CE
sin 75 / 10 = sin 90 / CE
CE = 10 sin 75 = BE ≈ 10.35
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