1281. Equilibrist squares

    Five acrobats perform balancing feats. They are the equilibrist squares... Now, they realize that the square A and the triangle A have the same area... Here's Nina Guindilla's proof:

    Dear Teacher:
    I may take the side of the square A as length unit, may I not? So, the area of the square A is 1, right? I have to prove that the area of the triangle A is 1 too. Look at the angles α, β, σ, τ, and sides s, t, below...

Area  =

=  s · t · sinβ / 2  =

=  s · t · sin(90°−σ−τ) / 2  =

=  s · t · cos(σ+τ) / 2  =

=  s · t · (cosσ · cosτ − sinσ · sinτ ) / 2  =

=  (1 · s · cosσ) · (1 · t · cosτ) / 2  −  (s · sinσ) · (t · sinτ) / 2  =

    I apply the law of cosines and the law of sines...

=  (1² + s² − sin²α) · (1² + t² − cos²α) / 8  −  sinα · sin(90°+α) · cosα · sin(180°−α) / 2  =

=  (1+1²−2·sinα·cos(90°+α))·(1+1²−2·cosα·cos(180°−α))/8 − sinα·cosα·cosα·sinα/2  =

=  (2 + 2 · sinα · sinα) · (2 + 2 · cosα · cosα) / 8  −  sin²α · cos²α / 2  =

=  (1 + 1 · sin²α) · (1 + 1 · cos²α) / 2  −  sin²α · cos²α / 2  =

=  (1 + sin²α + cos²α +  sin²α · cos²α −  sin²α · cos²α) / 2  =

=  (1 + 1 + 0) / 2  =

=  1