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 $BE$.

Rangcr897 Jun 17, 2024

#1**+1 **

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! :)

NotThatSmart Jun 17, 2024