Question

In $\Delta ABC,\angle ABC={135}^{\circ }$

Prove that:
$A{C}^2$=$A{B}^2$+ $B{C}^2$ + $4\left(\Delta ABC\right)$

Answer 1

Lets draw altitude from A to BC such that it meets BC at D, angles are $\Delta ADC,\angle D={90}^o$

$\Rightarrow A{D}^2+D{C}^2=A{C}^2$

$\Rightarrow A{D}^2+(BD+BC{)}^2=A{C}^2$

$\Rightarrow A{D}^2+B{D}^2+B{C}^2+2BD.BC=A{C}^2$ ………….$\left(1\right)$

In $\Delta ADB,\angle D={90}^o$

$\Rightarrow A{D}^2+B{D}^2=A{B}^2$

Substituting in equation $\left(1\right)$

$\Rightarrow A{D}^2+B{D}^2+B{C}^2+2BD.BC=A{C}^2$

$\Rightarrow A{B}^2+B{C}^2+2BD.BC=A{C}^2$

as in $\Delta ADB,\angle DAB=\angle {45}^o$

$\Rightarrow AD=DB$(opposite sides are equal)

$\Rightarrow A{B}^2+B{C}^2+2BD\times BC=A{C}^2$

$\Rightarrow A{B}^2+B{C}^2+2AD\times BC=A{C}^2$

$\Rightarrow A{B}^2+B{C}^2+4[\frac{1}{2}AD\times BC]=A{C}^2$

$\Rightarrow A{B}^2+B{C}^2+4\left(\Delta ABC\right)=A{C}^2$

$\therefore$ Hence proved.