ABCD is a cyclic quadrilateral and BC is a diameter of the related circle on which A and D also lie. $$\angle BCA = 19^{\circ}$$ and $$\angle CAD = 32^{\circ}$$. What is the measure of $$\angle ACD$$?

From the question, the following figure can be obtained. 

Since BC is a diameter, the angle subtended by it at the circumference is a right angle. $$\angle BAC=90^\circ. $$

Also, $$\angle BCA=19^\circ.$$

Using the angle sum property of a triangle, $$ \angle ABC =180^\circ-90^\circ-19^\circ =71^\circ.$$

Since (ABCD) is a cyclic quadrilateral, opposite angles are supplementary.

$$\angle ADC =180^\circ-\angle ABC =180^\circ-71^\circ =109^\circ.$$

Now consider $$\triangle ACD$$.

Given, $$\angle CAD=32^\circ. $$

Hence, $$\angle ACD =180^\circ-109^\circ-32^\circ =39^\circ. $$