SAT Question b1e1c2f5: Lines, Angles, and Triangles answer and explanation with Desmos solution

It's given that in triangle $ABC$, point $D$ on side $AB$ is connected by a line segment with point $E$ on side $AC$ such that line segment $DE$ is parallel to side $BC$. It follows that parallel segments $DE$ and $BC$ are intersected by sides $AB$ and $AC$. If two parallel segments are intersected by a third segment, corresponding angles are congruent. Thus, corresponding angles $C$ and $AED$ are congruent and corresponding angles $B$ and $ADE$ are congruent. Since triangle $ADE$ has two angles that are each congruent to an angle in triangle $ABC$, triangle $ADE$ is similar to triangle $ABC$ by the angle-angle similarity postulate, where side $DE$ corresponds to side $BC$, and side $AE$ corresponds to side $AC$. Since the lengths of corresponding sides in similar triangles are proportional, it follows that $\frac{DE}{BC}=\frac{AE}{AC}$. Since point $E$ lies on side $AC$, $AE+CE=AC$. It's given that $CE=2AE$. Substituting $2AE$ for $CE$ in the equation $AE+CE=AC$ yields $AE+2AE=AC$, or $3AE=AC$. It's given that $BC=162$. Substituting $162$ for $BC$ and $3AE$ for $AC$ in the equation $\frac{DE}{BC}=\frac{AE}{AC}$ yields $\frac{DE}{162}=\frac{AE}{3AE}$, or $\frac{DE}{162}=\frac{1}{3}$. Multiplying both sides of this equation by $162$ yields $DE=54$. Thus, the length of line segment $DE$ is $54$.