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

Answer


In right $\triangle\text{ALD},$
By Pythagoras theorem,
$AD^2 = AL^2 + DL^2$
$\Rightarrow AL^2 = AD^2 - DL^2 ....(i)$
In right $\triangle\text{ACL},$
By Pythagoras theorem,
$\text{AC}^2 =\text{ AL}^2 + \text{LC}^2$
$\Rightarrow \text{AC}^2 = (\text{AD}^2 -\text{ DL}^2) + (\text{DL} +\text{ DC)}^2\dots(\text{from (i))}$
$\Rightarrow\text{AC}^2=\text{AD}^2-\text{DL}^2+\Big(\text{DL}+\frac{\text{BC}}{2}\Big)^2$.....(Since AD is the median)
$\Rightarrow\text{AC}^2=\text{AD}^2-\text{DL}^2+\text{DL}^2+\Big(\frac{\text{BC}}{2}\Big)^2+\text{BC}.\text{DL}$
$\Rightarrow\text{AC}^2=\text{AD}^2+\text{BC}.\text{DL}+\Big(\frac{\text{BC}}{2}\Big)^2$
Hence proved.

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