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.

irrational, cannot