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.
Answer: Let m∠CLN = x. Then m∠ALM = 3x, and m∠A = 90°-x, m∠C = 90°-3x.
The sum of angles of ∆ABC is 180°, so we have
... 180° = 40° + m∠A + m∠C
Using the above expressions for m∠A and m∠C, we can write ...
... 180° = 40° + (90° -x) + (90° -3x)
... 4x = 40° . . . . . . . . . add 4x-180°
... x = 10°
From which we conclude ...
... m∠C = 90°-3x = 90° - 3·10° = 60°
The ratio of CN to CL is
... CN/CL = cos(∠C) = cos(60°)
... CN/CL = 1/2
so ...
... CN = (1/2)CL
