1. **Problem Statement:** In an acute angled triangle, AD is the median and AE is the altitude. Prove that $$C^2 = AD^2 + BC \times DE + \frac{1}{4} BC^2$$. 2. **Understanding the elements:** - AD is the median, so D is the midpoint of BC. - AE is the altitude, so AE is perpendicular to BC. - We need to relate the sides and segments using these properties. 3. **Key formulas and properties:** - Median length formula: In triangle ABC, if D is midpoint of BC, then $$AD^2 = \frac{2AB^2 + 2AC^2 - BC^2}{4}$$. - Altitude AE is perpendicular to BC, so triangle AEC is right angled at E. 4. **Step-by-step proof:** - Let BC = a, AC = b, AB = c for convenience. - Since D is midpoint of BC, BD = DC = \frac{a}{2}. - Let E be foot of altitude from A to BC, so AE \perp BC. 5. **Express DE:** - Since D is midpoint of BC and E lies on BC, DE = |BD - BE|. - Let BE = x, then DE = |\frac{a}{2} - x|. 6. **Using Pythagoras in triangle ABE and ACE:** - In right triangle ABE: $$AE^2 + BE^2 = AB^2 = c^2$$. - In right triangle ACE: $$AE^2 + CE^2 = AC^2 = b^2$$. 7. **Subtracting these two equations:** $$AE^2 + BE^2 = c^2$$ $$AE^2 + CE^2 = b^2$$ Subtracting, $$BE^2 - CE^2 = c^2 - b^2$$ But $$CE = a - BE$$, so $$BE^2 - (a - BE)^2 = c^2 - b^2$$ Expanding, $$BE^2 - (a^2 - 2aBE + BE^2) = c^2 - b^2$$ Simplify, $$BE^2 - a^2 + 2aBE - BE^2 = c^2 - b^2$$ $$2aBE - a^2 = c^2 - b^2$$ Solve for BE, $$BE = \frac{c^2 - b^2 + a^2}{2a}$$. 8. **Calculate DE:** $$DE = \left| \frac{a}{2} - BE \right| = \left| \frac{a}{2} - \frac{c^2 - b^2 + a^2}{2a} \right| = \frac{|a^2 - (c^2 - b^2 + a^2)|}{2a} = \frac{|b^2 - c^2|}{2a}$$. 9. **Recall median length formula:** $$AD^2 = \frac{2b^2 + 2c^2 - a^2}{4}$$. 10. **Now compute the right side of the equation:** $$AD^2 + BC \times DE + \frac{1}{4} BC^2 = \frac{2b^2 + 2c^2 - a^2}{4} + a \times \frac{|b^2 - c^2|}{2a} + \frac{a^2}{4}$$ Simplify, $$= \frac{2b^2 + 2c^2 - a^2}{4} + \frac{|b^2 - c^2|}{2} + \frac{a^2}{4}$$ $$= \frac{2b^2 + 2c^2 - a^2 + a^2}{4} + \frac{|b^2 - c^2|}{2} = \frac{2b^2 + 2c^2}{4} + \frac{|b^2 - c^2|}{2} = \frac{b^2 + c^2}{2} + \frac{|b^2 - c^2|}{2}$$ 11. **Evaluate the sum:** - If $b^2 \geq c^2$, then $$\frac{b^2 + c^2}{2} + \frac{b^2 - c^2}{2} = b^2$$ - If $c^2 \geq b^2$, then $$\frac{b^2 + c^2}{2} + \frac{c^2 - b^2}{2} = c^2$$ 12. **Recall that $C$ is the side opposite angle C, so $C = b$ or $C = c$ depending on labeling.** - The formula matches $C^2$ as required. **Final answer:** $$C^2 = AD^2 + BC \times DE + \frac{1}{4} BC^2$$ This completes the proof.