1. **State the problem:** We have points A, B, C, D on a circle, with C, D, E collinear. Given that $BA = BD$, $CB = CD$, and angle $ABD = 40^\circ$, we need to find the size of angle $ADE$. 2. **Identify key properties:** Since $BA = BD$, triangle $ABD$ is isosceles with $AB = BD$. Similarly, $CB = CD$ means triangle $BCD$ is isosceles with $BC = CD$. 3. **Use the isosceles triangle property:** In triangle $ABD$, angles opposite equal sides are equal. Since $BA = BD$, angles $ABD$ and $ADB$ are equal. Given angle $ABD = 40^\circ$, angle $ADB = 40^\circ$. 4. **Find angle $BAD$ in triangle $ABD$:** Sum of angles in triangle is $180^\circ$: $$\angle BAD + 40^\circ + 40^\circ = 180^\circ$$ $$\angle BAD = 180^\circ - 80^\circ = 100^\circ$$ 5. **Analyze triangle $BCD$:** Since $CB = CD$, triangle $BCD$ is isosceles with base $BD$. Angles opposite equal sides are equal, so angles $CBD$ and $CDB$ are equal. 6. **Use the fact that $A, B, C, D$ lie on a circle:** Opposite angles of cyclic quadrilateral sum to $180^\circ$. So, $$\angle BAD + \angle BCD = 180^\circ$$ We know $\angle BAD = 100^\circ$, so $$\angle BCD = 180^\circ - 100^\circ = 80^\circ$$ 7. **Find angles in triangle $BCD$:** Let $\angle CBD = \angle CDB = x$. Then, $$x + x + 80^\circ = 180^\circ$$ $$2x = 100^\circ$$ $$x = 50^\circ$$ 8. **Find angle $ADE$:** Since $C, D, E$ are collinear, angle $ADE$ is the exterior angle to triangle $ABD$ at vertex $D$. The exterior angle equals the sum of the two opposite interior angles: $$\angle ADE = \angle ABD + \angle BAD = 40^\circ + 100^\circ = 140^\circ$$ **Final answer:** $$\boxed{140^\circ}$$