Suppose that $ABCD$ is a trapezoid in which $\overline{AD}|| \overline{BC}$.  Given $\overline{AC}\perp\overline{CD}$, $\overline{AC}$ bisects angle $\angle BAD$, and $[ABCD]=42$, then compute $[\triangle ACD]$.
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 $\triangle BAC$ is isosceles.  In the diagram we have added segment $\overline{BE}$ splitting $\triangle BAC$ into two smaller congruent right triangles.  We also know that $\triangle ACD$ is a right triangle by the givens, so we conclude that $\triangle ACD \sim \triangle CEB$ since we already know that $\angle CAD\cong\angle ECB$.  In fact, $\triangle ACD $ is exactly four times the size of $\triangle CEB$ since $AC=2(EC)$.  If we let $[\triangle CEB]=K$, then $[\triangle AEB]=K$ while $[\triangle ACD]=4K$.  Thus $6K=42$, so $K=7$ and $[\triangle ACD]=4K=\boxed{28}$.

[asy]
import olympiad; import graph; size(150); defaultpen(linewidth(0.8)); dotfactor=4;
int randangle = 50;
draw((-5,0)--(5*dir(randangle))--(5,0)--cycle);
path x1 = (5*Cos(randangle),5*Sin(randangle))--(-10,5*Sin(randangle));
path x2 = (-5,0)--(5dir(2*randangle));
pair X = intersectionpoint(x1,x2);
draw((-5,0)--X--(5*dir(randangle)));
draw(rightanglemark((5,0),(5*dir(50)),(-5,0),s=14));
draw(anglemark((5,0),(-5,0),X,18));
draw(anglemark(X,5*dir(50),(-5,0),18));
label("$A$",(-5,0),W); label("$D$",(5,0),E); label("$C$",(5*dir(50)),E);
label("$B$",(X),N);

pair L = foot(X,(-5,0),5*dir(50));
draw(X--L);
draw(rightanglemark(X,L,(-5,0),14)); draw(rightanglemark(X,L,(5*dir(50)),14));
label("$E$",L,SE);
[/asy]