Let $M$ be the midpoint of $BC$. Let $B'$ and $C'$ be points on $AD$ such that $AB' = AB = 5$ and $DC' = DC = 6$. Then $B'C' = 17 - 5 - 6 = 6$. Note that $\triangle ABM \cong \triangle AB'M$ and $\triangle DCM \cong \triangle DC'M$. Note also that $\triangle MB'C'$ is isosceles as $MB' = MB = MC = MC'$. Let $N$ be the midpoint of $B'C'$. Then we have $B'N = C'N = 3$ and $\angle MB'C' = \angle MC'B'$. Observe that $MCDB'$ is cyclic since $MD$ bisects $\angle B'DC$ and $MB' = MC$ (while $DB' \ne DC$). Thus,
$$2 \angle B'MA = \angle B'MB = \angle C'DC = 2 \angle MDC'.$$

Finally we have $MN = \sqrt{MC'^2 - NC'^2} = \sqrt{30-9} = \sqrt{21}$. Therefore,
$$\begin{align*} [ABCD] &= 2([AB'M] + [DC'M] + [MB'N]) \\ &= (AB' + C'D + B'N) \times MN \\ &= (5+6+3)\sqrt{21} = 14\sqrt{21}. \end{align*}$$