Question

In $△ABC,\angle ABC={135}^o$. prove that
${AC}^2={AB}^2+{BC}^2+4A\left(△ABC\right)$

Answer 1

From cosine formula we know that

$\cos B=\frac{A{B}^2+B{C}^2-A{C}^2}{2AB.BC}$

$\cos {135}^0=\frac{A{B}^2+B{C}^2-A{C}^2}{2AB.BC}=\frac{-1}{\sqrt{2}}$

$⟹A{B}^2+B{C}^2-A{C}^2=-\sqrt{2}AB.BC$

$⟹A{C}^2=A{B}^2+B{C}^2+\sqrt{2}AB.BC$

Now, we know area of triangle $A\left(\Delta ABC\right)=\frac{1}{2}AB.BC\sin B$

$⟹A\left(\Delta ABC\right)=\frac{1}{2}AB.BC\sin {135}^o$

$⟹A\left(\Delta ABC\right)=\frac{1}{2}AB.BC\frac{1}{\sqrt{2}}$

$⟹AB.BC=2\sqrt{2}A\left(\Delta ABC\right)$

Hence

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

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