Inscribed Squares in Right-Angled Triangle

Theorem

For any right-angled triangle, two squares can be inscribed inside it.

One square would share a vertex with the right-angled vertex of the right-angled triangle:

The other square would have a side lying on the hypotenuse of the right-angled triangle:

Compass and Straightedge Construction

Both inscribed squares can be constructed using compass and straightedge, thus demonstrating their existence.

Side Lengths

Let $a, b, c$ be the side lengths of the right-angled triangle, where $c$ is the length of the hypotenuse.

Then the side lengths of the inscribed squares are given by:

Shared right angle

$l = \dfrac {a b} {a + b}$

Side lies on Hypotenuse

$l = \dfrac {a b c} {a b + c^2}$

Proof

By definition of inscribed polygon, all four vertices of the inscribed square lies on the sides of the right-angled triangle.

By Pigeonhole Principle, at least two of the vertices must lie on the same side of the right-angled triangle.

The case where this side is the hypotenuse would be the second case above.

For the case where this side is not the hypotenuse, if none of the vertices coincide with the right angle of the right-angled triangle, the angle formed by the three vertices not on the hypotenuse would be obtuse, which is not possible since a square contains no obtuse angles.

Therefore a vertex must coincide with the right angle of the right-angled triangle, which is the first case above.

$\blacksquare$