Brazilian Mathematical Olympiad

Let $M$ be the midpoint of $AC$, $N$ the midpoint of $BD$. Take $P$ so that $PM$ is parallel to $BD$ and $PN$ is parallel to $AC$. Now $EM$ is a midline of $\triangle ABC$, so $\text{area } AEM = \frac{\text{area } ABC}{4}$. Similarly, $\text{area } AHM = \frac{\text{area } ADC}{4}$. So $\text{area } MHAE = \frac{\text{area } ABCD}{4}$. But $MP$ is parallel to $EH$, so $\text{area } MEH = \text{area } PEH$. Hence

$$\begin{align*} \text{area } PHAE &= \text{area } PEH + \text{area } AEH \\ &= \text{area } MEH + \text{area } AEH \\ &= \text{area } MHAE = \frac{\text{area } ABCD}{4} \end{align*}$$

Similarly for $\text{area } PEBF$ and the other quadrilaterals.