+207 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 $BE$.
+1908 So we know that
\(CD = BD = 10 \\ ED = ED\)
Through this, we can find that triangle CDE is congruent to triangle BDE.
This means BE = CE!
\(sin CED / CD = sin CDE / CE \\ sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35\)
Thanks! :)
