Question

In △ABC, m∠ABC=40°,

BL
(L∈
AC
) is the angle bisector of ∠B. Point M∈
AB
so that
LM
⊥
AB
and N∈
BC
so that
LN
⊥
BC
.
b
Find the measures of the angles of △ABC if m∠CLN=3m∠ALM and prove that CN= 1 2 CL.

Answer 1

Answer: 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 2

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