1. **Problem statement:** We have an isosceles triangle $\triangle ABC$ with $AB = AC$ and $\angle A = 40^\circ$. Points $B$, $C$, and $D$ are collinear, and we want to find the measure of $\angle ACD$. 2. **Key properties:** In an isosceles triangle, the angles opposite the equal sides are equal. Since $AB = AC$, the angles at $B$ and $C$ are equal. 3. **Calculate angles at $B$ and $C$:** The sum of angles in a triangle is $180^\circ$. So, $$\angle A + \angle B + \angle C = 180^\circ$$ Given $\angle A = 40^\circ$ and $\angle B = \angle C$, let $\angle B = \angle C = x$. Then, $$40^\circ + x + x = 180^\circ$$ $$2x = 180^\circ - 40^\circ = 140^\circ$$ $$x = 70^\circ$$ 4. **Analyze points $B$, $C$, and $D$ collinear:** Since $B$, $C$, and $D$ lie on a straight line, the angle on a straight line is $180^\circ$. 5. **Find $\angle ACD$:** At point $C$, the straight line $BCD$ forms a $180^\circ$ angle. We know $\angle BCA = 70^\circ$ (from step 3). The angle $\angle ACD$ is the angle between $AC$ and $CD$, which is the supplementary angle to $\angle BCA$ on the line. Therefore, $$\angle ACD = 180^\circ - 70^\circ = 110^\circ$$ **Final answer:** $$\boxed{110^\circ}$$