1. **Problem statement:** In the figure, ABC is a straight line. E lies on BD produced such that AD = AE. Prove that $a + b = 180^\circ$. 2. **Formula and rules:** The sum of angles on a straight line is $180^\circ$. Also, in isosceles triangles, base angles are equal. 3. **Step-by-step solution:** 1. Given $AD = AE$, triangle $ADE$ is isosceles with $AD = AE$. 2. Therefore, angles opposite these equal sides are equal: $\angle DAE = \angle DEA = 2a$ (given $\angle EAD = 2a$). 3. Since $ABC$ is a straight line, $\angle DAB + \angle ABO = a + b$ lie on the straight line at point $B$. 4. By the property of angles on a straight line, $a + b = 180^\circ$. 4. **Explanation:** Because $AD = AE$, triangle $ADE$ is isosceles, so angles at $D$ and $E$ are equal. The angles $a$ and $b$ lie on a straight line, so their sum must be $180^\circ$. **Final answer:** $$a + b = 180^\circ$$