View attachment 12686
where [MATH]C = \theta + x[/MATH] and [MATH]c=e+d[/MATH]
[MATH]c^2=e^2+d^2+2de[/MATH]
[MATH]c^2 = (a^2-h^2) + (b^2 - h^2) + 2ed[/MATH]
[MATH]c^2 = a^2 + b^2 - 2h^2 + 2(a\sin\theta)(b\sin x)[/MATH]
[MATH]c^2 = a^2 + b^2 - 2h^2 + 2ab\sin\theta\sin x[/MATH] and using the cosine identity [MATH]\cos(\theta+x)=\cos\theta\cos x-\sin\theta\sin x[/MATH]
[MATH]c^2 = a^2 + b^2 - 2h^2 + 2ab\cos\theta\cos x - 2ab\cos(\theta+x)[/MATH]
[MATH]c^2 = a^2 + b^2 - 2h^2 + 2(a\cos\theta)(b\cos x) - 2ab\cos C[/MATH]
[MATH]c^2 = a^2 + b^2 - 2h^2 + 2h^2 - 2ab\cos(\theta+x)[/MATH]
[MATH]c^2 = a^2 + b^2 - 2ab\cos C[/MATH]
and for a proof for the [MATH]\cos(a+b)[/MATH] identity, the next image proves the sine identity(I took the proof from Philips Lloyd, not sure if someone else found it first)
View attachment 12687
and from that and the definition of sine and cosine:
[MATH]\sin(90 - x) = \cos x[/MATH] and vice versa
[MATH]\sin((90 - a) - b) = \sin(90 - a)\cos b - \cos(90 - a)\sin b[/MATH]
[MATH]\sin(90 - (a + b)) = \cos a\cos b - \sin a\sin b[/MATH]
[MATH]\cos(a+b) = \cos a\cos b - \sin a\sin b[/MATH]and anything before that is the area of triangles formula and definitions of the sine and cosine functions so yeah.