Question
In $\triangle\text{ABC},\text{D}$ is the midpoint of BC and $\text{AE}\perp\text{BC}.$ If $\text{AC}>\text{AB},$ show that.
$\text{AB}^2=\text{AD}^2-\text{BC}.\text{DE}+\frac{1}{4}\text{BC}^2.$

Answer

Given: $\triangle\text{ABC}$ in which D is the midpoint of BC.
$\text{AE}\perp\text{BC}$ and AC > AB.
Then BD = CD and $\angle\text{AED}=90^\circ,$
Then, $\angle\text{ADE}<90^\circ$ and $\angle\text{ADC}>90^\circ

In $\triangle\text{AED},$ $\angle\text{AED}=90^\circ$
$\therefore\text{AD}^2=\text{AE}^2+\text{DE}^2$
$\Rightarrow\text{AE}^2=\big(\text{AD}^2-\text{DE}^2\big)\dots(1)$
In $\triangle\text{AEB},\angle\text{AEB}=90^\circ$
$\therefore \text{AB}^2=\text{AE}^2+\text{BE}^2\dots(2)$
Putting value of $AE^2$ from (1) in (2),
we get $\therefore\text{AB}^2=\big(\text{AD}^2-\text{DE}^2\big)+\text{BE}^2$
$=\big(\text{AD}^2-\text{DE}^2\big)+\big(\text{BD}^2-\text{DE}^2\big)\Big[\text{But}\text{ BD}=\frac{1}{2}\text{BC}\Big]$
$=\text{AD}^2-\text{DE}^2+\Big(\frac{1}{2}\text{BC}-\text{DE}\Big)^2$
​​​​​​​ $=\text{AD}^2-\text{DE}^2+\frac{1}{4}\text{BC}^2+\text{DE}^2-\text{BC.DE}$ $\text{AB}^2=\text{AD}^2-\text{BC},\text{DE}+\frac{1}{4}\text{BC}^2$

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