In ΔABC, m∠ABC = 40°, BL is the angle bisector of ∠B with point L∈ AC. Point M ∈ AB so that LM ⊥ AB and N ∈ BC so that LN⊥ BCFind the angle measures in ΔABC if m∠CLN = 3m∠ALM and prove that CN = 1 /CL.

Check the picture below.M is perpendicular to AB and ∈ AB, and stemming from point L.N is perpendicular to BC and ∈ BC, and stemming from point L as well.BL is the bisector, that means the angle at verte B, gets cut into two equal halves.now, we know what ∠CLN =3∠ALM, namely that ∠CLN is 3 times greater than ∠ALM.so, once the bisector kicks in, you get two angles of 20° each, the angles at M are 90° each and the angles at N are 90° each as well, that pretty much narrows down what the missing angle is in triangles MBL and NBL, so is 70°.now, the little sliver angles at CLN and ALM are on a 3:1 ratio, so, the flat-line of AC affords us 180°, subtract the 140°, so CLN and ALM will have to share only the remaining 40°, and they have to do so on a 3:1 ratio, that leaves us with, notice the blue angles.