proof of the angle theorem

the angle theorem
For any triangle ΔABC, and any point P which is not on any
of the lines AB, BC, CA. Let <APB=γ, <BPC=α, <CPA=β, then
cos2(α) + cos2(β) + cos2(γ) = 2cos(α)cos(β)cos(γ) + 1.

The first part of the proof is visual. There are some figures on
the right to illustrate the idea.
If P is within ΔABC, then α+β+γ=2π.
If P is outside ΔABC, then one of the segments AP, BP,CP lies
within the angle formed by the other two. Then this angle is the
is the sum of the other two.

Note that, by our method of measuring angles α,β,γ lie in (0,π).

Recall the basic trigonometric fact that cos(θ) = cos(κ) if and
only if θ = κ+2kπ or θ = -κ+2kπ. Consider the equations
(a) cos(α+β) = cos(γ), and (b) cos(α-β) = cos(γ).
For (a) we require α+β = ±γ+2kπ.
But 0 < α+β < 2π, and 0 < γ < π so the only possibilities are
α+β = γ and α + β = 2π-γ, i.e. α+β+γ = 2π.
For (b) we require α-β = ±γ+2kπ.
We have -π < α-β < π, and 0 <γ < π, so the only possibilities are
α-β = γ, i.e. α = β+γ, and α-β=-γ, i.e. β = α+γ.

Comparing the two analyses, we see that we must have either
cos(α+β) = cos(γ) or cos(α-β) = cos(γ). These are easily seen
to be equivalent to the single condition :
(cos(α+β)-cos(γ))(cos(α-β)-cos(γ)) = 0.

Using standard trigonometric formulae, this becomes
(cos(α)cos(β)-cos(γ)-sin(α)sin(β))(cos(α)cos(β)-cos(γ)+sin(α)sin(β)) = 0.
Multiplying out, and using the fact that sin2(θ) = 1-cos2(θ), we get the
stated result.

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