Question
In a $\triangle\text{ABC},\text{AD}$ is a median and $\text{AL}\perp\text{BC}.$
Prove that:
Prove that:
- $\text{AC}^2=\text{AD}^2+\text{BC}.\text{DL}+\Big(\frac{\text{BC}}{2}\Big)^2$
- $\text{AB}^2=\text{AD}^2-\text{BC}.\text{DL}+\Big(\frac{\text{BC}}{2}\Big)^2$
- $\text{AC}^2+\text{AB}^2=2\text{AD}^2+\frac{\text{1}}{2}\text{BC}^2$


