1. **Problem Statement:** Prove that the altitude $AD$ of triangle $ABC$ is smaller than sides $AB$ and $AC$. Given $\angle BAC = 60^\circ$, $AC = 5$ cm, and $AD$ is the altitude from $A$ to $BC$. 2. **Relevant Formulas and Rules:** - The altitude $AD$ in a triangle is the perpendicular segment from a vertex to the opposite side. - In any triangle, the altitude is always shorter than the sides adjacent to the vertex from which it is drawn. - Using trigonometry, $AD = AC \times \sin(\angle BAC)$. 3. **Calculate the altitude $AD$:** Given $AC = 5$ cm and $\angle BAC = 60^\circ$, $$AD = AC \times \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \approx 4.33 \text{ cm}.$$ 4. **Compare $AD$ with $AC$ and $AB$:** - Since $AD \approx 4.33$ cm and $AC = 5$ cm, clearly $AD < AC$. - To find the possible values of $AB$, use the Law of Cosines: $$AB^2 = AC^2 + BC^2 - 2 \times AC \times BC \times \cos(\angle ACB).$$ However, without $BC$ or $\angle ACB$, we use the fact that $AD$ is the height from $A$ to $BC$, so $AD < AB$ because the altitude is always less than the side it is dropped from. 5. **Express $AB$ in terms of $AC$:** Since $AD = AC \sin(60^\circ)$ and $AD < AB$, then $$AB > AD = AC \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}.$$ Thus, the side $AB$ must be greater than $\frac{5\sqrt{3}}{2}$ cm. **Final conclusion:** $$AD < AB \quad \text{and} \quad AD < AC,$$ with $$AB > \frac{5\sqrt{3}}{2} \approx 4.33 \text{ cm}.$$