jmc

Since the measure of angle $BAC$ is 42, the other two equal angles of triangle $ABC$ must be $(180-42)/2=138/2=69$ degrees each. We will introduce the point $F$, which is the center of the circle, and draw segments from each vertex to $F$. Since segment $ED$ is tangent to the circle at point $C$, it must be perpendicular to radius $CF$. Angle $BAC$ is bisected by segment $AF$, so angle $FAC$ is 21 degrees. Angle $FCA$ is also 21 degrees since triangle $AFC$ is isosceles. Thus, the measure of angle $ACD$ is $90-21=\boxed{69\text{ degrees}}$, which is the same as the two base angles of triangle ABC.