1. **Problem statement:** In triangle $\triangle ABC$, medians $AD$ and $BE$ intersect at centroid $G$. Given $AB=10$, $AC=14$, and median $AD=8$, find the length of median $BE$. 2. **Recall the properties:** The centroid $G$ divides each median in the ratio $2:1$ starting from the vertex. 3. **Use Apollonius' theorem:** For any triangle, the length of a median can be found using the formula: $$m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}$$ where $m_a$ is the median from vertex $A$ to side $BC$, and $a$, $b$, $c$ are the sides opposite vertices $A$, $B$, and $C$ respectively. 4. **Assign sides:** - $a = BC$ (unknown) - $b = AC = 14$ - $c = AB = 10$ 5. **Use the given median $AD=8$:** $$8^2 = \frac{2(14^2) + 2(10^2) - a^2}{4}$$ $$64 = \frac{2(196) + 2(100) - a^2}{4}$$ $$64 = \frac{392 + 200 - a^2}{4}$$ $$64 = \frac{592 - a^2}{4}$$ Multiply both sides by 4: $$256 = 592 - a^2$$ Rearranged: $$a^2 = 592 - 256 = 336$$ $$a = \sqrt{336} = 4\sqrt{21}$$ 6. **Find median $BE$ (median from vertex $B$):** Using Apollonius' theorem for median $BE = m_b$: $$m_b^2 = \frac{2a^2 + 2c^2 - b^2}{4}$$ Substitute known values: $$m_b^2 = \frac{2(336) + 2(10^2) - 14^2}{4}$$ $$m_b^2 = \frac{672 + 200 - 196}{4} = \frac{676}{4} = 169$$ $$m_b = \sqrt{169} = 13$$ **Final answer:** The length of median $BE$ is $13$.