1. **Problem statement:** We have an equilateral triangle ABC, and side BC is extended to point E such that C is the midpoint of BE. We need to find the measures of angles $\angle ACE$ and $\angle AEC$. 2. **Given:** - Triangle ABC is equilateral, so all sides are equal: $AB = BC = CA$. - $C$ is the midpoint of $BE$, so $BC = CE$. 3. **Step 1: Understand the figure and lengths** Since $ABC$ is equilateral, $AB = BC = CA = s$ (say). Since $C$ is midpoint of $BE$, $BC = CE = s$. 4. **Step 2: Coordinates or vector approach (optional for clarity)** Place points for easier calculation: - Let $B$ be at origin $(0,0)$. - Since $BC = s$, and $C$ lies on the x-axis, $C$ is at $(s,0)$. - Since $C$ is midpoint of $BE$, $E$ is at $(2s,0)$. 5. **Step 3: Find coordinates of A** In equilateral triangle $ABC$, point $A$ lies above $BC$ at height $h = \frac{\sqrt{3}}{2}s$. So $A$ is at $(\frac{s}{2}, \frac{\sqrt{3}}{2}s)$. 6. **Step 4: Calculate vectors for angles** - Vector $CA = A - C = (\frac{s}{2} - s, \frac{\sqrt{3}}{2}s - 0) = (-\frac{s}{2}, \frac{\sqrt{3}}{2}s)$ - Vector $CE = E - C = (2s - s, 0 - 0) = (s, 0)$ 7. **Step 5: Calculate $\angle ACE$** Use dot product formula: $$\cos \theta = \frac{\vec{CA} \cdot \vec{CE}}{|\vec{CA}||\vec{CE}|}$$ Calculate dot product: $$\vec{CA} \cdot \vec{CE} = (-\frac{s}{2})(s) + (\frac{\sqrt{3}}{2}s)(0) = -\frac{s^2}{2}$$ Magnitudes: $$|\vec{CA}| = \sqrt{(-\frac{s}{2})^2 + (\frac{\sqrt{3}}{2}s)^2} = \sqrt{\frac{s^2}{4} + \frac{3s^2}{4}} = \sqrt{s^2} = s$$ $$|\vec{CE}| = |(s,0)| = s$$ So, $$\cos \theta = \frac{-\frac{s^2}{2}}{s \times s} = -\frac{1}{2}$$ Therefore, $$\theta = \cos^{-1}(-\frac{1}{2}) = 120^\circ$$ 8. **Step 6: Calculate $\angle AEC$** Vectors at point $E$: - Vector $EA = A - E = (\frac{s}{2} - 2s, \frac{\sqrt{3}}{2}s - 0) = (-\frac{3s}{2}, \frac{\sqrt{3}}{2}s)$ - Vector $EC = C - E = (s - 2s, 0 - 0) = (-s, 0)$ Dot product: $$\vec{EA} \cdot \vec{EC} = (-\frac{3s}{2})(-s) + (\frac{\sqrt{3}}{2}s)(0) = \frac{3s^2}{2}$$ Magnitudes: $$|\vec{EA}| = \sqrt{(-\frac{3s}{2})^2 + (\frac{\sqrt{3}}{2}s)^2} = \sqrt{\frac{9s^2}{4} + \frac{3s^2}{4}} = \sqrt{3s^2} = s\sqrt{3}$$ $$|\vec{EC}| = s$$ So, $$\cos \phi = \frac{\frac{3s^2}{2}}{s \times s\sqrt{3}} = \frac{3/2}{\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$$ Therefore, $$\phi = \cos^{-1}(\frac{\sqrt{3}}{2}) = 30^\circ$$ **Final answers:** - $\angle ACE = 120^\circ$ - $\angle AEC = 30^\circ$