+216 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 $BE$.
+1950 So we know that
CD=BD=10ED=ED
Through this, we can find that triangle CDE is congruent to triangle BDE.
This means BE = CE!
sinCED/CD=sinCDE/CEsin75/10=sin90/CECE=10sin75=BE≈10.35
Thanks! :)
