+339 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 graph the question.
From the graph and the question, we know that
CD=BD=10ED=ED∠CDE=∠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
sinCED/CD=sinCDE/CE
Plugging in some numbers, we find that
sin75/10=sin90/CECE=10sin75=BE≈10.35
Thus, 10.35 is approximately our answer.
Thanks! :)
*1800 points
