+253 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 $BE$.
+1897 Let's graph the question.
From the graph and the question, we know that
\(CD = BD = 10 \\ ED = ED \\ \angle CDE = \angle BDE = 90\)
So, by SAS congruence, we know that triangle CDE is congruent to triangle BDE
Thus, we can complete the problem in a series of steps using this congruence.
First, we know that
\(sin CED / CD = sin CDE / CE\)
Plugging in some numbers, we find that
\(sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35\)
Thus, 10.35 is approximately our answer.
Thanks! :)
*1800 points
