Angles Inside a Quadrilateral

Let the quadrilateral be $ABCD$. Using the provided angles, we can find that $∠BCA=90°$ and therefore point $C$ belongs to the circle with diameter $AB$. Also, we can see that $△AOD$ is isosceles, $OA=OD=OB$, and point $D$ also lies on the circle with diameter $AB$.

Using the provided angles, we can find that $∠BCA=90°$ and therefore point $C$ belongs to the circle with diameter $AB$. Also, we can see that $△AOD$ is isosceles, $OA=OD=OB$, and point $D$ also lies on the circle with diameter $AB$.

Therefore, $ABCD$ is inscribed in a circle with center $O$, and $∠ACD=∠AOD/2=40°$.