Question

If $D$ be the middle point of the side $BC$ of $△ABC$ whose area is $\Delta$ and $\angle ADB=\theta$, then prove $\frac{{AC}^2-{AB}^2}{4\Delta }=\cot \theta$

Answer 1

$BD=DC$

Let $AE\perp BC$

${AC}^2-{AB}^2=({AE}^2+{EC}^2)-({AE}^2+{BE}^2)$

$={EC}^2-{BE}^2=(EC+BE)(EC-BE)$

$=BC[(ED+DC)-(BD-ED)]=BC.2ED$

$[\therefore BD=DC]$ ...(1)

$4△=4\times \frac{1}{2}\times BC\times AE=2BC.AE$ ...(2)

From (1) is (2)

$\frac{{AC}^2-{AB}^2}{4△}=\frac{BC.2ED}{2BC.AE}=\frac{ED}{AE}=\cot \theta$