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Q. Prove that  ABC is right-angled at A if AB = 2n + 1, AC = 2n (n+ 1) and BC = 2n (n + 1) + 1.
 

Hi Vinay Don dear, being A as 90 degree
Hypotenuse BC = 2n(n+1) + 1
Sides AB = 2n + 1
And AC = 2n (n+1)
For the triangle to be right angled we have to prove
BC^2 = AB^2 + AC^2
Plugging on R H S
(2n+1)^2 + [2n (n+1)]^2 
==> [2n (n+1)]^2 + 4n^2 + 4n + 1
==> ​[2n (n+1)]^2 + 2* 2n (n + 1) + 1
==> A^2 + 2 * A * B + B^2 {here A = 2n(n+1), B = 1}
Thus we have (A+B)^2 
That is [2n(n+1) + 1}^2 = L H S
Thus by converse of Pythagoras it is confirmed that BAC is a right angled triangle 
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