In the adjoining figure, PL ⊥ QR; LQ = LS and LR > LQ. Show that PR > PQ.

In △PLQ and △PLS,

LQ = LS (Given)

PL = PL (Common)

∠PLQ = ∠PLS (Both are equal to 90°)

△PLQ ≅ △PLS (By S.A.S axiom)

We know that corresponding parts of congruent triangles are equal.

∠1 = ∠3

In △PRS,

∠3 > ∠2 (As exterior angle is greater than each interior opposite angle)

∴ ∠1 > ∠2

⇒ PR > PQ (As side opposite to greater angle is greater)

Hence, proved that PR > PQ.