Question

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

Answer 1

Given:

AB = 2n+1

AC = 2n(n+1)

BC = 2n(n+1)+1

To Prove:

ABC is right-angled at A

Solution:

Let us assume that angle A is 90 degrees

Then according to the Pythagorean theorem,

BC^2 = AC^2 + AB^2

(2n(n+1)+1)²  = (2n(n+1))² +(2n+1)²

Expanding RHS:

(2n(n+1))² +(2n+1)²

(2n(n+1))² + {4n²+4n+1}   --1

{4n²+4n+1} can be written as 2*2n(n+1)+1    --2

From 1 and 2,

2n(n+1) is common, thus we denote it as X and we denote 1 by Y   --3

Rewriting equation 1 using 3

X² + 2*X*Y +Y²   --4

Equation 4 can be rewritten as (X+Y)²

(X+Y)² is the same as LHS

Thus, by the converse of the Pythagoras theorem,  it is confirmed that BAC is a right-angled triangle.