In this figure $\angle RFS=\angle FDR$, $FD=4$ inches, $DR=6$ inches, $FR=5$ inches, $FS=7\dfrac{1}{2}$ inches. The length of $RS$, in inches, is:
$\textbf{(A) }\text{undetermined}\qquad\textbf{(B) }4\qquad\textbf{(C) }5\dfrac{1}{2}\qquad\textbf{(D) }6\qquad\textbf{(E) }6\dfrac{1}{2}\qquad$
We examine $\triangle RDF$ and $\triangle SFR$.  We are given $\angle RDF \cong \angle SFR$. Also note that $\frac{SF}{RD} = \frac{7.5}{6} = 1.25$ and $\frac{FR}{DF} = \frac{5}{4} = 1.25$, so $\frac{SF}{RD} = \frac{FR}{DF}$.
If two triangles have two pairs of sides that are proportional, and the included angles are congruent, then the two triangles are similar by SAS congruence.  Therefore, the third pair of sides must also be in the same proportion, so 
$\frac{SF}{RD} = \frac{FR}{DF} = \frac{RS}{FR} = 1.25$
$\frac{RS}{FR} = 1.25$
$\frac{RS}{5} = 1.25$
$RS = 6.25$, which is answer $\boxed{\textbf{(E) }}$