Given :

ΔPQR, AB ∥ QR

, and

To find: PB

In ΔPAB and ΔPQR

∠P = ∠P (Common)

(Corresponding angles as AB||QR with PQ as the transversal)

∠PAB = ∠PQR

(By AA similarity criteria)

⇒ ΔPAB ∼ ΔPQR

 (Corresponding Parts of Similar Triangles are propositional)

 AB/ QR = PB/ PR