18.16 In quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $O$, $OB=4$, $OD=6$, $OA=8$, $OC=3$, and $AB=6$. Then the length of $AD$ is 
(A) 9.
(B) 10.
(C) $6 \sqrt{3}$.
(D) $8 \sqrt{2}$.
(E) $\sqrt{166}$.
(18th American High School Mathematics Examination, 1967)
[Solution] Draw a perpendicular line $AF$ from $A$ to the extension of $DB$, and let $BF=x, AF=y$. Therefore, by the Pythagorean theorem, we have
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=6^{2} \\
(x+4)^{2}+y^{2}=8^{2}
\end{array}\right.
$$
Subtracting (1) from (2) gives $8 x+16=28$, which yields $x=\frac{3}{2}$.

Substituting $x$ into (1) gives $y^{2}=\frac{135}{4}$.
$$
\therefore A D^{2}=(10+x)^{2}+y^{2}=\left(\frac{23}{2}\right)^{2}+\frac{135}{4}=166 \text {. }
$$

Thus,
$$
A D=\sqrt{166} .
$$

Therefore, the answer is $(E)$.