1. **State the problem:** We have an equilateral triangle $\triangle ABC$ with points $D$ on $AB$ and $E$ on $AC$. Lines $BE$ and $CD$ intersect at $F$ such that $\angle BFC = 120^\circ$. We need to compare the lengths of $AD$ and $CE$. 2. **Understand the triangle properties:** Since $\triangle ABC$ is equilateral, all sides have equal length, say $s$. Also, $\angle ABC = \angle BCA = \angle CAB = 60^\circ$. 3. **Assign variables:** Let $AD = x$ and $CE = y$. Since $D$ is on $AB$, then $DB = AB - AD = s - x$. Similarly, $E$ is on $AC$ so $AE + EC = AC = s$, thus $AE = s - y$. 4. **Parameterize points:** - Point $D$ divides $AB$ at fraction $\frac{x}{s}$ along $AB$ from $A$ to $B$. - Point $E$ divides $AC$ at fraction $\frac{s - y}{s}$ along $AC$ from $A$ to $C$. 5. **Coordinates:** Let's place $\triangle ABC$ in coordinate plane: - $A = (0, \sqrt{3}/2 s)$ (top vertex) - $B = (-s/2, 0)$ - $C = (s/2, 0)$ Then, - $D = A + \frac{x}{s}(B - A) = \left(-\frac{x}{2}, \sqrt{3}/2 s - \frac{\sqrt{3}}{2} x \right)$ - $E = A + \frac{s - y}{s}(C - A) = \left(\frac{s - y}{2}, \sqrt{3}/2 s - \frac{\sqrt{3}}{2} (s - y) \right)$ 6. **Write parametric equations of lines:** - Line $BE$: from $B(-s/2,0)$ to $E \left(\frac{s - y}{2}, \sqrt{3}/2 s - \frac{\sqrt{3}}{2} (s - y) \right)$ - Line $CD$: from $C(s/2,0)$ to $D \left(-\frac{x}{2}, \sqrt{3}/2 s - \frac{\sqrt{3}}{2} x \right)$ 7. **Find intersection point $F$ of $BE$ and $CD$:** - Set parametric equations equal: $B + t (E - B) = C + u (D - C)$. Solving for $t$ and $u$ we get expressions for $F$ in terms of $x, y, s$. 8. **Compute vectors $\overrightarrow{FB}$ and $\overrightarrow{FC}$:** Using $F$, $B$, and $C$, find these vectors. 9. **Find angle between vectors $FB$ and $FC$:** The angle $\theta$ between vectors $\mathbf{u}$ and $\mathbf{v}$ is $$\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}.$$ Given $\theta = 120^\circ$, $\cos 120^\circ = -\frac{1}{2}$. From this, set up an equation relating $x$ and $y$. 10. **Solve the resulting equation:** After simplification, it follows that the only way for the intersection angle to be $120^\circ$ is if $$x = y,$$ meaning the lengths $AD$ and $CE$ are equal. **Answer:** (C) $AD = CE$