1. Let $O$ be the circumcenter of the acute triangle $ABC$. This means $O$ is the point equidistant from all three vertices $A$, $B$, and $C$ of the triangle.
2. Point $P$ is in the interior of triangle $AOB$. This implies that $P$ lies within the region bounded by the segments $AO$, $BO$, and $AB$.
3. Let $D$, $E$, and $F$ be the projections of $P$ on the sides $BC$, $CA$, and $AB$, respectively. This means $D$, $E$, and $F$ are the feet of the perpendiculars dropped from $P$ to the respective sides.
4. We need to prove that the parallelogram formed by $FE$ and $FD$ as its adjacent sides lies inside triangle $ABC$.

To prove this, we will show that the angles formed by the sides of the parallelogram with the sides of the triangle are such that the parallelogram remains within the triangle.

5. Let $\angle BAC = \alpha$, $\angle ABC = \beta$, and $\angle ACB = \gamma$.
6. Let $K$ be the fourth vertex of the parallelogram $FEKD$. By the properties of a parallelogram, $K$ is determined such that $FE \parallel DK$ and $FD \parallel EK$.

7. We need to show that $\angle AEF + \angle FEK < 180^\circ$ and $\angle BDF + \angle FDK < 180^\circ$.

8. Consider the angle $\angle AEF$. Since $E$ and $F$ are projections, $\angle AEF$ is related to the angles at $P$ and the angles of triangle $ABC$.

9. We have:
$$90^\circ + \angle OBP > 90^\circ$$
This implies:
$$\gamma + \angle PAF + \angle OAP + \angle OBP > 90^\circ$$
Therefore:
$$\angle EAP + \angle PBD = \gamma + \angle OAP + \angle OBP > 90^\circ - \angle PAF = \angle APF = \angle AEF$$
This leads to:
$$\angle EFD > \angle AEF$$
Hence:
$$\angle AEF + 180^\circ - \angle EFD < 180^\circ$$
Which simplifies to:
$$\angle AEF + \angle FEK < 180^\circ$$

10. Similarly, we can show that:
$$\angle BDF + \angle FDK < 180^\circ$$

11. Since both conditions are satisfied, the parallelogram $FEKD$ lies inside triangle $ABC$.

$\blacksquare$