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 $BE = \boxed{10 \sqrt{2}}$.
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 $BE$.
1 answer