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

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- Jul 19th 2010, 11:21 PMGGGGGGGGGGGGGGGGGGReally need help on this hard problem IMO
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 - Jul 19th 2010, 11:46 PMundefined
- Jul 20th 2010, 01:26 AMCaptainBlack
- Jul 20th 2010, 02:45 AMArchie Meade
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

$\displaystyle 2|OA|^2=|AC|^2\ \Rightarrow\ |OA|=\frac{|AC|}{\sqrt{2}}\ \Rightarrow\ 2|OA|=\sqrt{2}|AC|$