Area of irregular quadrilateral with diagonals proof

Question

I have an irregular convex quadrilateral with diagonals d and D. These diagonals form an acute angle $\alpha$. I know that I can find the area of this quadrilateral by using this formula:

$A = \frac{D\cdot d}{2} \sin(\alpha)$

But what is the proof for this formula? I've already tried using sine's law and cross product, but I could find the proof.

Answer 1

Since $\bf D$ and $\bf d$ intersect, suppose the point of intersection cuts $\bf D$ to segments of length $x, y$, and $\bf d$ to $z,w$.

The sine of the angles formed between $\bf d$ and $\bf D$ are equal, acute or obtuse.

Using the formula $Area = \frac12bc\sin A$:

$$Area = \frac12xz\sin\alpha+\frac12zy\sin\alpha+\frac12yw\sin\alpha+\frac12wx\sin\alpha = \frac12(w+z)(x+y)\sin\alpha = \frac{Dd}2\sin\alpha$$