Prove that, ‘In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the angle opposite to the first side is a right angle’.

Answer: Proof of the converse of Pythagoras theorem

[Figure: \(\Delta ABC\) with \(AC^2 = AB^2 + BC^2\); a second triangle PQR with \(\angle Q = 90^\circ\), PQ = AB and QR = BC.]

Given: In \(\Delta ABC\), \(AC^2 = AB^2 + BC^2\). To prove: \(\angle ABC = 90^\circ\).

Construction: Draw \(\Delta PQR\) such that \(\angle PQR = 90^\circ\), PQ = AB and QR = BC.

Proof: In \(\Delta PQR\), \(\angle Q = 90^\circ\), so by Pythagoras theorem \(PR^2 = PQ^2 + QR^2 = AB^2 + BC^2\) ...... (I)

Given \(AC^2 = AB^2 + BC^2\) ...... (II). From (I) and (II), \(PR^2 = AC^2\) \(\therefore PR = AC\).

In \(\Delta ABC\) and \(\Delta PQR\): AB = PQ, BC = QR, AC = PR. \(\therefore \Delta ABC \cong \Delta PQR\) (SSS test).

\(\therefore \angle ABC \cong \angle PQR = 90^\circ\) (c.a.c.t.). Hence proved.