In the diagram below, a triangle is drawn within a rectangle:

By considering the area of the rectangle, prove that

1. $$ad-bc=pq\sin \theta$$

2. Hence, by considering ${\sin }^2\theta +{\cos }^2\theta =1$, prove that

   $$ac+bd=pq\cos \theta$$

It is clear that the area of the rectangle is equal to $ad.$

However, we can also compute the area by summing the four triangles:

Equating these should lead to part (a).

For part (b), rearrange part (a) to make $\sin \theta$ the subject and square both sides.

Now use the fact that ${\sin }^2\theta +{\cos }^2\theta =1$ to replace ${\sin }^2\theta$ with $1-{\cos }^2\theta .$

There is quite a bit of algebra to do, but stick with it: it does work out in the end.

(At some point, you will need to use $p^2=a^2+b^2$ and $q^2=c^2+d^2.$)