1. **Problem Statement:** Given a triangle $\triangle ABC$ with $AB = BC$ and $\angle ABC = 60^\circ$, find the area of $\triangle ABC$. 2. **Known Information:** - $AB = BC$ means $\triangle ABC$ is isosceles with $AB = BC$. - $\angle ABC = 60^\circ$. 3. **Formula for Area of a Triangle:** The area of a triangle with two sides $a$ and $b$ and included angle $\theta$ is given by: $$\text{Area} = \frac{1}{2}ab \sin \theta$$ 4. **Apply to $\triangle ABC$:** Since $AB = BC$, let $AB = BC = x$. The included angle between these sides at $B$ is $60^\circ$. 5. **Calculate Area:** $$\text{Area} = \frac{1}{2} \times x \times x \times \sin 60^\circ = \frac{1}{2} x^2 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} x^2$$ 6. **Final Answer:** The area of $\triangle ABC$ is: $$\boxed{\frac{\sqrt{3}}{4} x^2}$$ where $x = AB = BC$. This formula gives the area in terms of the side length $x$.