Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$, and $\overline{AD}\perp\overline{BD}$. Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and let $P$ be the midpoint of $\overline{BD}$. Given that $OP=11$, the length of $AD$ can be written in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$?
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$
Angle chasing* reveals that $\triangle BPC\sim\triangle BDA$, therefore 
\[2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43},\]
or $AB=86$.
Additional angle chasing shows that $\triangle ABO\sim\triangle CDO$, therefore
\[2=\frac{AB}{CD}=\frac{BO}{OD}=\frac{BP+11}{BP-11},\]
or $BP=33$ and $BD=66$.
Since $\triangle ADB$ is right, the Pythagorean theorem implies that
\[AD=\sqrt{86^2-66^2}=4\sqrt{190}.\]
The answer is $4+190=\boxed{\textbf{(D) }194}$.

Angle Chasing: If we set $\angle DBC = \alpha$, then we know that $\angle DCB = 180^\circ-2\alpha$ because $\triangle DBC$ is isosceles. Then, $\angle BCP = 90^\circ-\alpha$, so $\angle BPC$ is a right angle. Because $\angle BDC = \alpha$ and $\overline{AB}\parallel\overline{DC}$, we conclude that $\angle ABD = \alpha$ too. Lastly, because $\triangle BPC$ and $\triangle BDA$ are both right triangles, they are similar by AA.
~mn28407 (Solution)
~mm (Angle Chasing Remark)
~eagleye ~MRENTHUSIASM ~charyyu83 (Minor Edits)