1. **Problem Statement:** Given a quadrilateral with points A, B, C, D where sides AB = BC = CD, find the measure of angle $\angle ABC$. 2. **Key Information:** Since AB = BC = CD, triangle ABC and triangle BCD share equal sides, indicating isosceles triangles. 3. **Step 1: Understand the equal sides** - AB = BC = CD means $\triangle ABC$ and $\triangle BCD$ have two sides equal. 4. **Step 2: Use the given angles** - Angle at point F inside the figure on BC is 115°. - Angle at point D is 40°. 5. **Step 3: Analyze triangle BCD** - Since BC = CD and angle at D is 40°, the base angles at B and C in $\triangle BCD$ are equal. - Sum of angles in $\triangle BCD$ is 180°, so $$\angle B + \angle C + 40^\circ = 180^\circ$$ $$2\angle B + 40^\circ = 180^\circ$$ $$2\angle B = 140^\circ$$ $$\angle B = 70^\circ$$ 6. **Step 4: Find $\angle ABC$** - $\angle ABC$ is the angle at B in $\triangle ABC$. - Since AB = BC, $\triangle ABC$ is isosceles with base AC. - The angle at B in $\triangle ABC$ is the vertex angle. 7. **Step 5: Use the straight line property** - The angle at B in $\triangle BCD$ is 70°, and the angle at B in $\triangle ABC$ plus this 70° must sum to 115° (angle at F on line BC). - So, $$\angle ABC + 70^\circ = 115^\circ$$ $$\angle ABC = 115^\circ - 70^\circ = 45^\circ$$ **Final answer:** $$\boxed{45^\circ}$$