1. **State the problem:** We need to find the size of angle $c$ in a figure with two triangles sharing a middle segment, given angles $20^\circ$ and $62^\circ$, and two equal-length sides on the right. 2. **Analyze the figure:** The top line has a vertex near the middle, forming two triangles. The left top angle is $20^\circ$, the lower middle angle is $62^\circ$, and the two right segments are equal in length, indicating an isosceles triangle on the right. 3. **Use triangle angle sum rule:** The sum of angles in any triangle is $180^\circ$. 4. **Focus on the right triangle:** Since the two right segments are equal, the base angles opposite these sides are equal. Let each base angle be $c$. 5. **Calculate the third angle in the right triangle:** The angle adjacent to $c$ is $62^\circ$ (given). The sum of angles in this triangle is: $$ 62^\circ + c + c = 180^\circ $$ 6. **Simplify and solve for $c$:** $$ 62^\circ + 2c = 180^\circ $$ $$ 2c = 180^\circ - 62^\circ $$ $$ 2c = 118^\circ $$ $$ c = \frac{118^\circ}{2} $$ $$ c = 59^\circ $$ 7. **Final answer:** The size of angle $c$ is $59^\circ$.