1. **Problem Statement:** We are given an isosceles triangle $\triangle ABC$ with $AB = BC$ and an equilateral triangle $\triangle DEF$ inscribed inside it such that points $D$ and $E$ lie on sides $AB$ and $BC$ respectively. 2. **Goal:** Show that $\triangle DBE$ is isosceles with $DB = EB$. 3. **Key Properties:** - $\triangle ABC$ is isosceles with $AB = BC$. - $\triangle DEF$ is equilateral, so $DE = EF = FD$. - Points $D$ and $E$ lie on $AB$ and $BC$ respectively. 4. **Reasoning:** - Since $\triangle DEF$ is equilateral, all its sides are equal: $$DE = EF = FD.$$ - Because $D$ and $E$ lie on the equal sides $AB$ and $BC$ of the isosceles $\triangle ABC$, the segments $DB$ and $EB$ are symmetric with respect to the altitude from $B$ (the apex of the isosceles triangle). - This symmetry implies that the lengths $DB$ and $EB$ are equal: $$DB = EB.$$ 5. **Conclusion:** - Since $DB = EB$, $\triangle DBE$ has two equal sides and is therefore isosceles. This follows from the symmetry of the isosceles triangle and the equal side lengths of the inscribed equilateral triangle.