1. Given: Rectangle $ABCD$ is inscribed in circle $O$. Let the projections of a point $P$ on minor arc $CD$ onto $AB, AC, BD$ be $K, L, M$, respectively. We need to prove that $\angle LKM = 45^\circ$ if and only if $ABCD$ is a square.

2. Concyclic Points: Since $P$ is on the minor arc $CD$, and $K, L, M$ are projections of $P$ onto $AB, AC, BD$ respectively, the points $PAKL, PBKM, PLOM$ are concyclic. This is because the projections of a point onto the sides of a rectangle inscribed in a circle form right angles with the sides, making the quadrilaterals cyclic.

3. Angle Calculation:
$$\angle LKM = \angle PKL + \angle PKM$$
Since $PAKL$ and $PBKM$ are cyclic quadrilaterals, we have:
$$\angle PKL = \angle PAL \quad \text{and} \quad \angle PKM = \angle PBM$$
Therefore,
$$\angle LKM = \angle PAL + \angle PBM$$

4. Central Angles:
$$\angle PAL \quad \text{and} \quad \angle PBM$$
are half of the central angles subtended by the arcs $PL$ and $PM$ respectively. Let $\angle POL$ and $\angle POM$ be the central angles subtended by arcs $PL$ and $PM$:
$$\angle PAL = \frac{1}{2} \angle POL \quad \text{and} \quad \angle PBM = \frac{1}{2} \angle POM$$
Thus,
$$\angle LKM = \frac{1}{2} (\angle POL + \angle POM)$$

5. **Angle $\angle LOM$**:
Since $L$ and $M$ are projections of $P$ onto $AC$ and $BD$, $\angle LOM$ is the angle subtended by the arc $LM$ at the center $O$. Therefore,
$$\angle LKM = \frac{1}{2} \angle LOM$$

6. **Condition for $\angle LKM = 45^\circ$**:
For $\angle LKM = 45^\circ$,
$$\frac{1}{2} \angle LOM = 45^\circ \implies \angle LOM = 90^\circ$$

7. **Rectangle $ABCD$ is a Square**:
The angle $\angle LOM = 90^\circ$ implies that the diagonals of the rectangle $ABCD$ are perpendicular. This is a property of a square. Therefore, $ABCD$ must be a square if $\angle LOM = 90^\circ$.

8. Conclusion:
$$\angle LKM = 45^\circ \iff \angle LOM = 90^\circ \iff ABCD \text{ is a square}$$

$\blacksquare$