In the given figure, D is the midpoint of side BC and $\text{AE}\perp\text{BC}.$ If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that.
$(\text{b}^2-\text{c}^2)=2\text{ax}$
$(\text{b}^2-\text{c}^2)=2\text{ax}$
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Given: D is the midpoint of side $\text{BC},\text{AE}\perp\text{BC},\text{BC}=\text{a},\text{AC}=\text{b},\text{AB}=\text{c},\text{ED}=\text{x},\text{AD}=\text{p}$ and $\text{AE}=\text{h}$
In $\triangle\text{AEC},\angle\text{AEC}=90^\circ$
$\text{AD}^2=\text{AE}^2+\text{ED}^2$ (by pythagoras theorem)
$\Rightarrow\text{p}^2=\text{h}^2+\text{x}^2$
(iv) Subtracting (2) from (1), we get
$\text{b}^2-\text{c}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}-\bigg(\text{p}^2-\text{ax}+\frac{\text{a}^2}{4}\bigg)$
$=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}-\text{p}^2+\text{ax}-\frac{\text{a}^2}{4}$
$\big(\text{b}^2-\text{c}^2\big)=2\text{ax}$
In $\triangle\text{AEC},\angle\text{AEC}=90^\circ$
$\text{AD}^2=\text{AE}^2+\text{ED}^2$ (by pythagoras theorem)
$\Rightarrow\text{p}^2=\text{h}^2+\text{x}^2$
(iv) Subtracting (2) from (1), we get
$\text{b}^2-\text{c}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}-\bigg(\text{p}^2-\text{ax}+\frac{\text{a}^2}{4}\bigg)$
$=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}-\text{p}^2+\text{ax}-\frac{\text{a}^2}{4}$
$\big(\text{b}^2-\text{c}^2\big)=2\text{ax}$
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