ABCD is a cyclic quadrilateral and BC is a diameter of the circle on which points A and D also lie. Given that $\angle BCA = 19^\circ$ and $\angle CAD = 32^\circ$, find the measure of $\angle ACD$.

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**Solution:**

1. Since BC is a diameter of the circle, the angles subtended by BC at points A and D on the circle are right angles:
   - $\angle BAC = 90^\circ$
   - $\angle BDC = 90^\circ$

2. In triangle $ABC$, the sum of angles is $180^\circ$:
   $$\angle ABC + \angle BCA + \angle BAC = 180^\circ$$
   Substitute known values:
   $$\angle ABC + 19^\circ + 90^\circ = 180^\circ$$
   $$\angle ABC = 180^\circ - 109^\circ = 71^\circ$$

3. The angle $\angle ABC$ is composed of two angles at point B:
   $$\angle ABC = \angle ABD + \angle CBD$$

4. Given $\angle CAD = 32^\circ$, and since $\angle CAD$ and $\angle ABD$ subtend the same arc $CD$ in the circle, they are equal:
   $$\angle ABD = 32^\circ$$

5. Therefore, from step 3:
   $$\angle CBD = \angle ABC - \angle ABD = 71^\circ - 32^\circ = 39^\circ$$

6. Angles $\angle ACD$ and $\angle CBD$ subtend the same arc $BD$, so they are equal:
   $$\angle ACD = \angle CBD = 39^\circ$$

**Answer:** $39^\circ$

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