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 $BE$.
Let $F$ be the foot of the altitude from $D$ to $AC$. [asy]
unitsize(2 cm);
pair A, B, C, D, E, F;
B = (0,0);
C = (2,0);
A = extension(B, B + dir(75), C, C + dir(180 - 60));
E = extension(B, reflect(B,C)*(B + dir(75)), A, C);
D = extension(B, midpoint(B--C), A, E);
F = (D + E)/2;
draw(A--B--C--cycle);
draw(B--D);
draw(F--D);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, SE);
label("$E$", E, NE);
label("$F$", F, N);
[/asy]
Since $\angle C = 15^\circ$, $\angle A = 180^\circ - \angle B - \angle C = 180^\circ - (75^\circ + 15^\circ) = 90^\circ$. Therefore, triangle $ABC$ is a 45-45-90 triangle, so $AC = BC = 20$, and $AEC$ is a 30-60-90 triangle.
Since $AD$ and $BE$ are perpendiculars to $AC$, $BAD$ and $BCE$ are also 45-45-90 triangles. Therefore, $AD = FE = \frac{20}{\sqrt{2}} = 10 \sqrt{2}$.
Since $D$ is the midpoint of $\overline{BC}$, triangle $BDC$ is also a 45-45-90 triangle. Therefore, $BD = \frac{BC}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10 \sqrt{2}$.
Therefore, by SAS Congruence, triangle $BCE$ is congruent to triangle $FAD$, so $BE = \boxed{10 \sqrt{2}}$.
To find the length of $BE$, we can use the fact that the perpendicular bisector of a segment passes through the midpoint of the segment.
Let's start by finding the length of $BD$. Since $D$ is the midpoint of $BC$, we know that $BD = CD$.
Since $\angle C = 15^\circ$, then $\angle B = 180^\circ - \angle A - \angle C = 180^\circ - \angle A - 15^\circ = \angle A - 165^\circ$.
Since $\angle C = \angle DBC$ (by definition of perpendicular bisector), we have $\angle ADB = \angle BDC = \angle A - 165^\circ$.
Since $\angle A + \angle B + \angle C = 180^\circ$, we have $\angle A + (\angle A - 165^\circ) + 15^\circ = 180^\circ$.
Simplifying this equation gives $2\angle A = 330^\circ$, so $\angle A = 165^\circ$.
Since $\angle A = \angle ADB$, we have $BD = \frac{BC}{2} = \frac{20}{2} = 10$.
Since $D$ is the midpoint of $BC$, we have $BD = CD = 10$.
Now, let's use the fact that the perpendicular bisector of $BC$ passes through the midpoint of the segment $AC$. Let $F$ be the midpoint of $AC$.
Since $D$ is the midpoint of $BC$ and $F$ is the midpoint of $AC$, we have $DF \parallel BE$.
Since $DF \parallel BE$ and $BD = CD$, we have $BE = EF$.
Therefore, the length of $BE$ is equal to the length of $EF$.
To find the length of $EF$, we can use the fact that the triangles $ABC$ and $DEF$ are similar.
Since $\angle A = \angle DFE$, $\angle B = \angle DEF$, and $\angle C = \angle EFD$, the triangles $ABC$ and $DEF$ are similar by Angle-Angle Similarity.
Since $\frac{BD}{DF} = \frac{BC}{AC} = \frac{20}{AC}$ (by similarity), we have $\frac{10}{EF} = \frac{20}{AC}$.
Simplifying this equation gives $EF = \frac{10}{2} = 5$.
Therefore, the length of $BE$ is $\boxed{5}$.
To find the length of $BE$ in triangle $ABC$, we will use the properties of perpendicular bisectors and the information given.
First, let's draw triangle $ABC$ and construct the perpendicular bisector of $BC$. Since the perpendicular bisector of $BC$ passes through the midpoint of $BC$, let's mark the midpoint of $BC$ as point $M$.
Next, because $DM$ is the perpendicular bisector of $BC$, $DM$ is perpendicular to $BC$. Therefore, we can mark $\angle BDM$ as $90^\circ$.
We are given that $BC = 20$, so the distance from $B$ to $M$ along the perpendicular bisector of $BC$ is half of $BC$ which is $10$.
Since $\angle C = 15^\circ$, we know that $\angle MCB = \frac{1}{2} \angle C = 7.5^\circ$.
Finally, we can use trigonometry to find the length of $BE$. In triangle $BME$, we have the opposite side ($BE$) and the adjacent side ($BM$), and we want to find the length of $BE$.
We can use the tangent function, which is defined as opposite over adjacent, to find $BE$. Hence, we have:
$$\tan(7.5^\circ) = \frac{BE}{10}.$$
Solving for $BE$, we have:
$$BE = 10 \cdot \tan(7.5^\circ).$$
Using a calculator, we find that $\tan(7.5^\circ) \approx 0.1317$. Therefore, the length of $BE$ is approximately:
$$BE \approx 10 \cdot 0.1317 = 1.317.$$