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 $y + x = 11,$ $y = 2x,$ $2x + x = 11,$ $x = 11/3.$

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 $y/x$ = AB/CD. We note that $y=11-x$ and AB = 2CD, so that AB/CD = 2 to rewrite the equation as $(11 – x)/x = 2$. Solving for $x$, we get, again, that $x = 11/3$.

The answer, once again, is (a).