Iranian Geometry Olympiad

To solve the problem, we need to find all quadrilaterals $ABCD$ such that the four triangles $DAB$, $CDA$, $BCD$, and $ABC$ are similar to one another. Let's follow the logical steps to reach the conclusion.

1. Understanding Similarity of Triangles:

Triangles are similar if they have the same set of angles, or equivalently, if their corresponding sides are in proportional ratios. For the problem, we have:
- $\triangle$ DAB $\sim \triangle$ CDA $\sim \triangle$ BCD $\sim \triangle$ ABC.

2. Analyzing Angle Conditions:

Since all four triangles are similar, this gives us angle conditions:
- $\angle DAB = \angle CDA = \angle BCD = \angle ABC$.
- $\angle DBC = \angle BAC = \angle ACD = \angle ABD$.
- $\angle ADC = \angle BDA = \angle CBA = \angle DCA$.

3. Using Quadrilateral Properties:

In quadrilateral $ABCD$, the sum of interior angles is $360^\circ$. Combining with the angle conditions from similarities, let's explore:

Let $\angle DAB = \angle ABC = x$, $\angle ABC = \angle CDA = y$, and $\angle CDA = \angle BCD = z$.

Then since the sum of angles in each triangle is $180^\circ$, we have:
$$x + y + z = 180^\circ.$$

4. Conclusion with Rectangles:

Suppose $ABCD$ is a rectangle:
- Each angle in a rectangle is $90^\circ$.
- Therefore, each of $\triangle DAB, \triangle CDA, \triangle BCD, \triangle ABC$ has angles $90^\circ, 45^\circ, 45^\circ$.
- Triangles with these angles are all similar to each other as they preserve the angle conditions.

5. Verifying that Only Rectangles Work:

For a quadrilateral to have all internal triangles similar, only rectangles can satisfy these angle relations because they exactly distribute $360^\circ$ into four sets of equal pairs of angles.

Thus, we conclude that the quadrilaterals satisfying the given condition are:

$$\boxed{\text{All rectangles.}}$$