Question

In Figure, $ABC$ is a triangle in which $\angle ABC<{90}^{\circ }$ and $AD\perp BC.$ Prove that $A{C}^2=A{B}^2+B{C}^2-2BC.DB.$

Answer 1

Using Pythagoras Theorem in $\Delta ADB$ and $\Delta ADC$
In $\Delta ADB$
On applying Pythagoras theorem,
$A{B}^2=A{D}^2+D{B}^2.....\left(1\right)$
In $\Delta ADC$
On applying Pythagoras theorem,
$A{C}^2=A{D}^2+D{C}^2$
$A{C}^2=A{D}^2+(BC-BD{)}^2$
$A{C}^2=A{D}^2+B{C}^2+B{D}^2-2\left(BC\right)\left(DB\right)$
From (1)
$\Rightarrow A{C}^2=A{B}^2+B{C}^2-2\left(BC\right)\left(DB\right)$
Hence Proved.