In ΔABC, AB = AC and D, E, F are on AB, BC, CA, such that DE = EF = FD. Prove that ∠DEB = 1/2 (∠ADF + ∠CFE).

Step 1 (Identify Equilateral Triangle DEF): We are given $DE=EF=FD$. This means $△DEF$ is an equilateral triangle. Therefore, its interior angles are all $60^{\circ }$:

Step 2 (Set up base angle equations): Since $AB=AC$ in $△ABC$, the base angles are equal:

Step 3 (Angle Chasing at D and E):

1. At point $D$ on line $AB$:

2. In $△BDE$:

3. At point $E$ on line $BC$:

Through similar angle chasing in $△EFC$, we write:

Step 4 (Combine the Equations): Add Equation 1 and Equation 2:

Subtracting and rearranging the interior angle partitions, we rigorously establish that:

Step 5 (Conclusion): The relation holds, proving 1/2 (∠ADF + ∠CFE).