Cross's theorem

Proof by rotation

Rotate ${\displaystyle \triangle AID}$ by a right angle, such that ${\displaystyle D}$ coincides with ${\displaystyle B}$, and let this new triangle be ${\displaystyle \triangle AI'B}$. It is clear that ${\displaystyle |AC|=|AI'|}$, and that ${\displaystyle C}$, ${\displaystyle A}$, and ${\displaystyle I}$ are collinear. Therefore, the areas of triangles ${\displaystyle \triangle ABC}$ and ${\displaystyle \triangle AI'B}$ are equal. Since ${\displaystyle \triangle AI'B}$ is simply a rotation of ${\displaystyle \triangle AID}$, it follows that ${\displaystyle \triangle ABC}$ and ${\displaystyle \triangle AID}$ have the same area. Similar arguments prove the equality of all four areas.

Proof by formula for area

Observe that ${\displaystyle \angle CAB}$ and ${\displaystyle \angle DAI}$ are supplementary. Therefore, we have

${\displaystyle {\begin{aligned}[][DAI]&={\frac {1}{2}}|AD||AI|\sin(\angle DAI)\\&={\frac {1}{2}}|AB||AC|\sin(\angle CAB)\\&=[ABC],\end{aligned}}}$

as desired. Similar arguments show all four areas are equal.