If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude, 
then the ratio of the smaller base to the larger base is: 
$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{3}{4} \qquad \textbf{(D)}\ \frac{3}{5}\qquad \textbf{(E)}\ \frac{2}{5}$
Let $a$ be the length of the smaller base of isosceles trapezoid $ABCD$, and $1$ be the length of the larger base of the trapezoid. The ratio of the smaller base to the larger base is $\frac a1=a$. Let point $E$ be the foot of the altitude from $C$ to $\overline{AD}$.
Since the larger base of the isosceles trapezoid equals a diagonal, $AC=AD=1$. Since the smaller base equals the altitude, $BC=CE=a$. Since the trapezoid is isosceles, $DE=\frac{1-a}{2}$, so $AE = 1-\frac{1-a}{2} = \frac{a+1}{2}$. Using the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) on right triangle $ACE$, we have
\[a^2+\left(\frac{a+1}{2}\right)^2=1\]
Multiplying both sides by $4$ gives
\[4a^2+(a+1)^2=4\]
Expanding the squared binomial and rearranging gives
\[5a^2+2a-3=0\]
This can be factored into $(5a-3)(a+1)=0$. Since a must be positive, $5a+3=0$, so $a=\frac 35$. The ratio of the smaller base to the larger base is $\boxed{\textbf{(D) } \frac 35}$.