3580.11 – Trapezoid

Referring to the figure, part (I), we know $AC = 11$; we want $EC$ which we call $x$. Extend $AD$ and $BC$ to meet at $F$, forming $\triangle AFB$. Because $\triangle FDC \sim \triangle FAB$ and $AB = 2DC$, we know $FA = 2FD$ and $FD = DA$. Also, $FC = CB$. Thus $AC$ and $BD$ are medians, intersecting each other in the ratio $2:1$. So $y + x = 11$, $y = 2x$, $2x + x = 11$, and $x = 11/3.$

The answer is (a).

Alternatively, consider triangles $\triangle ABE$ and $\triangle CDE$ as in the figure, part (II). Since $CD$ and $AB$ are parallel, $\angle ABE \cong \angle CDE$, and $\angle EAB \cong \angle ECD$. Lines $AC$ and $BD$ intersect at point $E$, so $\angle DEC \cong \angle BEA$.

Since the angles of the two triangles are the same measure, we can say $\triangle ABE \sim \triangle CDE$, which means that $\dfrac{y}{x} = \dfrac{AB}{CD}$. We note that $y=11 - x$ and $AB = 2CD$, so that $\dfrac{AB}{CD} = 2$ to rewrite the equation as $\dfrac{11 – x}{x} = 2$. Solving for $x$, we get, again, that $x = \dfrac{11}{3}$.

The answer, once again, is (a).