In \(\Delta PQR\), \(\angle PQR\) is an obtuse angle. Seg PS \(\perp\) line QR, with R – Q – S. Prove that \(PR^2 = PQ^2 + QR^2 + 2\,QR \times QS\).

Answer: \(PR^2 = PQ^2 + QR^2 + 2\,QR \times QS\) (proved below)

[Figure: \(\Delta PQR\) with obtuse angle at Q; RQ is extended beyond Q to S so that PS is perpendicular to line RS.]

R – Q – S, so \(RS = RQ + QS\).

In right-angled \(\Delta PSR\): \(PR^2 = PS^2 + RS^2 = PS^2 + (QR + QS)^2 = PS^2 + QR^2 + 2\,QR \times QS + QS^2\) ...... (I)

In right-angled \(\Delta PSQ\): \(PS^2 + QS^2 = PQ^2\) ...... (II)

Substituting (II) in (I): \(PR^2 = PQ^2 + QR^2 + 2\,QR \times QS\). Hence proved.

Note: this shows that the side opposite an obtuse angle is longer than what Pythagoras theorem alone would give.