1. **Problem statement:** Calculate the size of angle $BCD$ given that angle $ABC$ is $128^\circ$ and the triangle has vertices $A$, $B$, $C$, and $D$ with $AB$ vertical and $AD$ horizontal. 2. **Understanding the problem:** We know angle $ABC = 128^\circ$. Since $AB$ is vertical and $AD$ is horizontal, angle $BAD$ is a right angle, i.e., $90^\circ$. 3. **Key fact:** The sum of angles around point $B$ on a straight line is $180^\circ$. Since $AB$ is vertical and $AD$ is horizontal, angle $ABC$ and angle $BCD$ are related through the geometry of the figure. 4. **Calculate angle $ABC$ relative to $BAD$:** Since $BAD = 90^\circ$ and $ABC = 128^\circ$, angle $ABC$ extends $38^\circ$ beyond the straight line $BAD$ because $128^\circ - 90^\circ = 38^\circ$. 5. **Calculate angle $BCD$:** Angle $BCD$ is the external angle to triangle $BCD$ at vertex $C$. By the exterior angle theorem, angle $BCD$ equals the sum of the two opposite interior angles. Since $BCD$ is marked in purple and relates to the given $128^\circ$ angle, we find that angle $BCD = 180^\circ - 128^\circ = 52^\circ$. 6. **Final answer:** The size of angle $BCD$ is $52^\circ$.