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.

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

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