SAT Question dba6a25a: Right Triangles And Trigonometry answer and explanation with Desmos solution

In the figure given, since $\overline{BD}$ is parallel to $\overline{AE}$ and both segments are intersected by $\overline{CE}$, then angle $BDC$ and angle $AEC$ are corresponding angles and therefore congruent. Angle $BCD$ and angle $ACE$ are also congruent because they are the same angle. Triangle $BCD$ and triangle $ACE$ are similar because if two angles of one triangle are congruent to two angles of another triangle, the triangles are similar. Since triangle $BCD$ and triangle $ACE$ are similar, their corresponding sides are proportional. So in triangle $BCD$ and triangle $ACE$, $\overline{BD}$ corresponds to $\overline{AE}$ and $\overline{CD}$ corresponds to $\overline{CE}$. Therefore, $\frac{\text{the length of }\overline{BD}}{\text{the length of }\overline{CD}}=\frac{\text{the length of }\overline{AE}}{\text{the length of }\overline{CE}}$. Since triangle $BCD$ is a right triangle, the Pythagorean theorem can be used to give the value of $CD$: $6^2+8^2=C{D}^2$. Taking the square root of each side gives $CD=10$. Substituting the values in the proportion yields $\frac{6}{10}=\frac{18}{CE}$. Multiplying each side by $CE$, and then multiplying by $\frac{10}{6}$ yields $CE=30$. Therefore, the length of $\overline{CE}$ is $30$.