In $\triangle ABC$, $\angle A = 90^\circ$, $AC = 3$ and $AB=4$. $CD$ is the angle bisector of $\angle ACB$. Point $E$ on $AC$ is such that $CE=AE$. $CD$ and $BE$ intersect at $M$. If the area of quadrilateral $ADME= \dfrac ab$, where $a$ and $b$ are positive coprime integers, find $a+b$.