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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 $BE$.

 Nov 17, 2024
 #1
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Let's graph (yay!)

 

What's interesting about this is this is on a coordinate plane, so we can make quite a few deductions from it. 

First off, we have that

\(CD = BD = 10 \\ ED = ED\)

 

Also though, note that ED and BC are just the x and y axis, thus they are perpendicular lines. 

Which means that we have \(\angle CDE = \angle BDE\)

 

By SAS Congruence, we know that triangle CDE is congruent to triangle BDE.

Thus, we can conclude that corresponding sides \(CE = BE\)

 

Finally, we just simply right that

\(sin CED / CD = sin CDE / CE \\ sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35\)

 

Thus, 10.35 SHOULD be correct, not sure, trig is never my thing :(

 

Thanks! :)

 Nov 18, 2024
edited by NotThatSmart  Nov 18, 2024

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