1. **Problem Statement:** We are given that in triangle ABC, the sides and angles satisfy the relation $$a^2 (1 + \cos A) = 2bc \sin^2 A$$, and we need to determine the shape of the triangle ABC. 2. **Recall the Law of Cosines:** For triangle ABC, where side $a$ is opposite angle $A$, the Law of Cosines states: $$a^2 = b^2 + c^2 - 2bc \cos A$$ 3. **Analyze the given relation:** Start from the given equation: $$a^2 (1 + \cos A) = 2bc \sin^2 A$$ Using the Pythagorean identity, rewrite $$\sin^2 A = 1 - \cos^2 A$$. Substitute this: $$a^2 (1 + \cos A) = 2bc (1 - \cos^2 A)$$ 4. **Rewrite and simplify:** $$a^2 (1 + \cos A) = 2bc (1 - \cos A)(1 + \cos A)$$ Because $$(1 - \cos^2 A) = (1 - \cos A)(1 + \cos A)$$ as a difference of squares. 5. **Divide both sides by $(1 + \cos A)$:** Provided that $1 + \cos A \neq 0$, divide both sides: $$a^2 = 2bc (1 - \cos A)$$ 6. **Compare with Law of Cosines:** From step 2, $$a^2 = b^2 + c^2 - 2bc \cos A$$ From step 5, $$a^2 = 2bc - 2bc \cos A$$ 7. **Set them equal:** $$b^2 + c^2 - 2bc \cos A = 2bc - 2bc \cos A$$ Simplify by adding $2bc \cos A$ on both sides and subtracting: $$b^2 + c^2 = 2bc$$ 8. **Interpret this equality:** The equation $$b^2 + c^2 = 2bc$$ is the same as $$b^2 - 2bc + c^2 = 0$$ which factors as $$(b - c)^2 = 0$$ which implies $$b = c$$ 9. **Conclusion:** Since sides $b$ and $c$ are equal, triangle ABC is **isosceles** with $b = c$. **Final answer:** The triangle ABC is an isosceles triangle.