Let ABC be a triangle such that angle ACB = 135°. Prove that:
$\displaystyle AB^2=AC^2+BC^2+\sqrt{2}\times AC\times BC$
I really have no Idea how to solve this, any help is really appreciated. Thanks
Let ABC be a triangle such that angle ACB = 135°. Prove that:
$\displaystyle AB^2=AC^2+BC^2+\sqrt{2}\times AC\times BC$
I really have no Idea how to solve this, any help is really appreciated. Thanks
You can also form a right-angled triangle by standing the triangle on side CB (or side AC).
Drop a vertical to O from A and join O to C such that OCB is a straight line.
This requires only Pythagoras' Theorem.
Then..
$\displaystyle |OA|^2+|OC|^2=|AC|^2$
$\displaystyle |OA|^2+\left(|OC|+|CB|\right)^2=|AB|^2$
$\displaystyle |OA|=|OC|$ since $\displaystyle |\angle OCA|=45^o$
$\displaystyle |OA|^2+|OA|^2+2|OA||CB|+|CB|^2=|AB|^2$
Finish with
$\displaystyle 2|OA|^2=|AC|^2\ \Rightarrow\ |OA|=\frac{|AC|}{\sqrt{2}}\ \Rightarrow\ 2|OA|=\sqrt{2}|AC|$