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Solve this:

Q. Prove that $\u2206$ *ABC* is right-angled at *A* if *AB *= 2*n* + 1, *AC* = 2*n* (*n*+ 1) and* BC* = 2*n* (*n* + 1) + 1.

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