1. **Problem statement:** Given a circle with center $C$, points $A$, $B$, and $D$ on the circumference, and $\angle ABD = 38^\circ$, find the measure of $\angle CAD$ in triangle $ACD$. 2. **Key facts and formulas:** - $C$ is the center of the circle, so $CA$ and $CD$ are radii and thus equal. - Triangle $ACD$ is isosceles with $CA = CD$. - The angle at the center ($\angle ACD$) subtends the arc $AD$. - The angle at the circumference ($\angle ABD$) subtending the same arc $AD$ is half the angle at the center. 3. **Find $\angle ACD$:** Since $\angle ABD = 38^\circ$ is an inscribed angle subtending arc $AD$, the central angle $\angle ACD$ is twice that: $$\angle ACD = 2 \times 38^\circ = 76^\circ$$ 4. **Triangle $ACD$ properties:** Since $CA = CD$, triangle $ACD$ is isosceles with base $AD$. 5. **Find $\angle CAD$:** Sum of angles in triangle $ACD$ is $180^\circ$: $$\angle CAD + \angle CDA + \angle ACD = 180^\circ$$ Since $\angle CAD = \angle CDA$ (isosceles triangle), let each be $x$: $$x + x + 76^\circ = 180^\circ$$ $$2x = 180^\circ - 76^\circ = 104^\circ$$ $$x = 52^\circ$$ **Final answer:** $$\boxed{\angle CAD = 52^\circ}$$