In $\triangle\text{ABC},\text{D}$ is the midpoint of $BC$ and $\text{AE}\perp\text{BC}.$ If $\text{AC}>\text{AB},$ show that.
$\text{AB}^2=\text{AD}^2-\text{BC}.\text{DE}+\frac{1}{4}\text{BC}^2.$
$\text{AB}^2=\text{AD}^2-\text{BC}.\text{DE}+\frac{1}{4}\text{BC}^2.$
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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 $\therefore\text{AB}^2=\big(\text{AD}^2-\text{DE}^2\big)+\text{BE}^2$ $=\big(\text{AD}^2-\text{DE}^2\big)+\big(\text{BD}^2-\text{DE}^2\big)\Big[\text{But}\text{ BD}=\frac{1}{2}\text{BC}\Big]$ $=\text{AD}^2-\text{DE}^2+\Big(\frac{1}{2}\text{BC}-\text{DE}\Big)^2$ $=\text{AD}^2-\text{DE}^2+\frac{1}{4}\text{BC}^2+\text{DE}^2-\text{BC.DE}$ $\text{AB}^2=\text{AD}^2-\text{BC},\text{DE}+\frac{1}{4}\text{BC}^2$
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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