Question

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⊥ BC

Find the angle measures in ΔABC if m∠CLN = 3m∠ALM and prove that CN = 1 /CL.

Answer 1

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

Answer 2

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.