Question

Prove that if in a triangle the square of one side is equal to the sum of the squares of the
other two sides, than the angle opposite to the first side is a right angle.
OR

Answer 1

Let there exists a triangle ABC such that,

$\mathsf{AC^2=AB^2+BC^2\quad\longrightarrow\quad(1)}$

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But by law of cosines, we have,

$\mathsf{AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot\cos\angle ABC}\quad\longrightarrow\quad(2)$

Then, from (1) and (2),

$\mathsf{AB\cdot BC\cdot\cos\angle ABC=0}$

Since $\mathsf{AB,\ BC\neq0,}$

$\mathsf{\cos\angle ABC=0}$

This implies,

$\mathsf{\angle ABC=90^{\circ}}$

Hence Proved!