Question

In $△ABC,\angle ABC={135}^o$. Prove that $A{C}^2=A{B}^2+B{C}^2+4A\left(△ABC\right)$.

Answer 1

R.E.F. Image.

so from the figure if it is

clear that AD = DB

Area of $△ABC=\frac{1}{2}BC\times AD=\frac{1}{2}BC\times DB$

so $BC\times DB=2ar\left(△ABC\right)$

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

$A{C}^2=A{D}^2+(DB+BC{)}^2$

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

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