In the adjoining figure, ABCD is a parallelogram in which AB = 16 cm, BC = 10 cm and L is a point on AC such that CL : LA = 2 : 3. If BL produced meets CD at M and AD produced at N, prove that :

(i) ΔCLB ∼ ΔALN.

(ii) ΔCLM ∼ ΔALB.

(i) In ΔCLB and ΔALN,

∠CBL = ∠ANL [Alternate angles are equal]

∠CLB = ∠ALN [Vertically opposite angles are equal]

∴ ΔCLB ∼ ΔALN (By A.A. axiom)

Hence, proved that ΔCLB ∼ ΔALN.

(ii) In ΔCLM and ΔALB,

∠LMC = ∠LAB [Alternate angles, since CM ∥ AB, transversal LM]

∠CLM = ∠ALB [Vertically opposite angles are equal]

∴ ΔCLM ∼ ΔALB (By A.A. axiom)

Hence, proved that ΔCLM ∼ ΔALB.