jmc

There are many ways to draw the diagram; one possibility is shown below. We know that $\angle BAC\cong \angle CAD\cong \angle BCA$ since $\overline{AC}$ bisects $\angle BAD$ and $\overline{AD}||\overline{BC}$. Therefore $△BAC$ is isosceles. In the diagram we have added segment $\overline{BE}$ splitting $△BAC$ into two smaller congruent right triangles. We also know that $△ACD$ is a right triangle by the givens, so we conclude that $△ACD\sim △CEB$ since we already know that $\angle CAD\cong \angle ECB$. In fact, $△ACD$ is exactly four times the size of $△CEB$ since $AC=2(EC)$. If we let $[△CEB]=K$, then $[△AEB]=K$ while $[△ACD]=4K$. Thus $6K=42$, so $K=7$ and $[△ACD]=4K=\boxed{28}$.