Proving Lami’s Theorem
Consider three forces P, Q, and R exerting over a single point O. Let the angles opposite to these forces be α, β, and γ respectively.
By considering the lines of forces of P and Q, let’s denote P = OA and Q = OB and complete the parallelogram OACB having OA and OB as its adjacent sides.
As these forces must be in equilibrium, their resultant must fall in line with OD and should be equal to R, however in the opposite direction.
Also the resultant of the forces P and Q will be expressed by the diagonal OC, through magnitude as well as direction of the parallelogram OACB.
The geometry of the diagram suggests,
BC = P and AC = Q
Therefore, ∟AOC = (180^o – β)
And ∟ACO = ∟BOC = (180^o – α)
Therefore ∟CAO = 180 – (∟AOC + ∟ACO)
= 180^o – [(180^o – β) + (180^o – α)]
= 180^o – 180^o + β – 180^o + α = α + β – 180^o, but since α + β+ γ = 360^o, we substract 180^o from both the sides and get,
(α + β – 180^o) + γ = 360^o + 180^o or ∟CAO = (180^o – γ),
Since for triangle AOC,
OA/sin (180^o – α ) = AC/sin (180^o – β) = OC/sin (180^o – γ), we finally get,
P/sin α = Q/sin β = R/sin γ, because sin (180^o – Ɵ) = sin Ɵ.
The last expression agrees with Lami’s theorem.