Converse Of Pythagoras Theorem

Theorem: in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.

Prerequisites:
Pythagoras theorem (proof)
SSS congruence (proof)

Proof:

Let there be a $\triangle ABC$, such that:

$\qquad\quad AB^2 + BC^2 = AC^2$

We need to prove that $\triangle ABC$ is a right triangle. For this, we construct a right triangle $\triangle PQR$, right angled at Q, such that PQ $=$ AB and QR $=$ BC.

Since, $\triangle PQR$ is a right triangle, hence by using pythagoras theorem, we get:

$\qquad\quad PQ^2 + QR^2 = PR^2\\
\Rightarrow\quad\;\; AB^2 + BC^2 = PR^2\qquad\qquad\qquad\text{(by construction)}\qquad\qquad\!\cdots\text{(1)}$
But, $\quad AB^2 + BC^2 = AC^2\qquad\qquad\qquad\text{(given)}\qquad\qquad\qquad\qquad\;\cdots\text{(2)}$

From $(1)$ and $(2)$,

$\qquad\quad PR^2 = AC^2\\
\Rightarrow\quad\;\; PR = AC$

Also, since PQ $=$ AB and QR $=$ BC by construction, hence by SSS congruency rule,

$\qquad\quad\triangle ABC\cong\triangle PQR$
$\Rightarrow\quad\;\;\angle B = \angle Q$
But, $\quad\angle Q = 90^o$
$\therefore\quad\;\;\;\angle B = 90^o$

Thus, $\triangle ABC$ is a right triangle, right angled at B.

Q.E.D.

Recommended;
Median through hypotenuse
Basic Proportionality Theorem
Location of centroid