Prove (µ+λ) cot θ = µ cot α - λ cot β?

Question

By refering to the diagram below,
http://img295.imageshack.us/my.php?image=untitlediv4.jpg

where D divides BC in the ratio µ : λ
Prove that (µ+λ) cot θ = µ cot α - λ cot β

Answer 1

Use the Law of Sines for this one. From inspection of the diagram, we have the following two equations:

Sin(θ-α)/AB = Sin(α)/µBC
Sin(180-θ-β)/AB = Sin(β)/λBC

Dividing one by the other, and getting rid of that 180 degrees, and rearranging, we have:

µSin(θ-α)Sin(β) - λSin(θ+β)Sin(α) = 0

Expanding those Sines and dividing everything by
Sin(θ)Sin(β)Sin(α), we end up with

(µ+λ)Cot(θ) - λCot(β) + µCot(α) = 0

Answer 2

(µ+λ) cot θ = µ cot α - λ cot β
X2
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= GET SOME FRIENDS AND GO HAVE A DRINK.

Answer 3

coming