+972 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 Let's make some observations about the problem. Note that
CD=BD=10ED=ED
From this, we can identify that triangle CDE and triangle BDE are congruent triangles.
This infromation tells us that BE=CE
Now, using the this new information, we can write the equation
sinCED/CD=sinCDE/CEsin75/10=sin90/CECE=10sin75=BE≈10.35
So our answer is about 10.35
Thanks! :)
