+948 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 $BE$.
+1926 Let's make some observations about the problem. Note that
\(CD = BD = 10 \\ ED = ED\)
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
\(sin CED / CD = sin CDE / CE \\ sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35\)
So our answer is about 10.35
Thanks! :)
