Given:- Right angle Triangle
To Prove:- Square of the hypotenuse is equal to the sum of the squares of the other two sides
Let ΔAOB be right angle triangle with right angle at O
![](https://www.sarthaks.com/?qa=blob&qa_blobid=14239356871719747562)
Thus we have to prove
AB2 = OA2 + OB2
Proof: - Let, O at Origin, then
\(\vec a\) and \(\vec b\) be position vector of A and B respectively
Since OB is perpendicular at OA, their dot product equals to zero
We know that,
![](https://www.sarthaks.com/?qa=blob&qa_blobid=5986369338311456069)
Now, We can see that, by triangle law of vector addition, \(\vec{AB}=\vec b-\vec a\)
Therefore,
![](https://www.sarthaks.com/?qa=blob&qa_blobid=11336954917195441831)
⇒ AB2 = OA2 + OB2(Pythagoras theorem)
Hence, proved.