SAT Question 1dacfb94: Lines, Angles, and Triangles answer and explanation with Desmos solution

It's given that $AB=BC$, so triangle $ABC$ is isosceles. It's also given that line segments $DE$ and $AC$ are parallel. Since line segments $DE$ and $AC$ are parallel, triangle $BDE$ is similar to triangle $BAC$, which means triangle $BDE$ is also isosceles. It’s given that point $O$ is the midpoint of line segment $DE$. Since triangle $BDE$ is isosceles, line segment $BO$ splits triangle $BDE$ into two congruent triangles, where $\angle BOD$ is a right angle. Therefore, $\angle BFA$ is also a right angle. The figure shows that the measure of $\angle OAF$ is $30^{\circ }$. It follows that triangle $AOF$ is a 30-60-90 triangle, where the measure of $\angle AOF$ is $60^{\circ }$. In a 30-60-90 triangle, the side opposite the $30^{\circ }$ angle is half the length of the side opposite the $90^{\circ }$ angle. Let $x$ represent the length of line segment $OF$ and let $2x$ represent the length of line segment $AO$. It's given that $AO=BO$. Since $AO=BO$, the length of line segment $BO$ is also $2x$. It follows that the length of line segment $BF$ is $BO+OF$, which is $2x+x$, or $3x$. Since triangle $BDE$ is similar to triangle $BAC$, the ratio of corresponding side lengths is equal to the ratio of the heights of the triangles. This means that $\frac{BE}{BC}=\frac{BO}{BF}$. Substituting $2x$ for $BO$ and $3x$ for $BF$ in this equation yields $\frac{BE}{BC}=\frac{2x}{3x}$, or $\frac{BE}{BC}=\frac{2}{3}$. It’s given that $AB=81$ and $AB=BC$, so $BC=81$. Substituting $81$ for $BC$ in the equation $\frac{BE}{BC}=\frac{2}{3}$ yields $\frac{BE}{81}=\frac{2}{3}$. Multiplying both sides of this equation by $81$ yields $BE=\frac{2}{3}(81)$, or $BE=54$.