Question

If a $△ABC,\angle C$ is an acute angle. $AE\perp BC$. Prove that $A{B}^2=A{D}^2-BC\times DE+\frac{B{C}^2}{4}$

Answer 1

We have,

$ADE$ is a right angled triangle.

Therefore ,By Pythagoras theorem,

$A{E}^2+E{D}^2=A{D}^2$

$\Rightarrow A{E}^2=A{D}^2-E{D}^2......\left(1\right)$

Again, $In\Delta AEB$

$A{B}^2=A{E}^2+B{E}^2$ ……. (2)

And now,

$BE=BD-ED$ …… (3)

Substituting equation (2) and (3) in (1), we get,

$A{B}^2=A{D}^2-E{D}^2+{(BD-ED)}^2$

$=A{D}^2-E{D}^2+B{D}^2+E{D}^2-2BD\times ED$

Now, $BD=\frac{BC}{2}(\therefore Dis\text{midpoint})$

Therefore,

$A{B}^2=A{D}^2+B{D}^2-2BD\times ED$

$=A{D}^2+{\left(\frac{BC}{2}\right)}^2-2\left(\frac{BC}{2}\right)\times ED$

$=A{D}^2-BC\times ED+\frac{B{C}^2}{4}$

Hence, this is the answer.