Inscribed Squares in Right-Angled Triangle/Compass and Straightedge Construction/Shared Right Angle

Theorem

Let $\triangle ABC$ be a right-angled triangle, where $\angle C = 90^\circ$.

Construct the angle bisector of $\angle C$.

Let the point of intersection of this angle bisector and side $AB$ be $E$.

Construct perpedicular lines from $E$ to sides $AC$ and $BC$, and name their intersections $F$ and $D$ respectively.

Then $CDEF$ is an inscribed square.

Proof

Note that $\angle C = 90^\circ$ and $ED \perp BC$, $EF \perp AC$ by construction.

Therefore $CDEF$ is a rectangle.

By definition of an angle bisector, $\angle ECF = 45^\circ$.

Since $\angle CFE = 90^\circ$, by Sum of Angles of Triangle equals Two Right Angles:

$\angle CEF + \angle CFE + \angle ECF = 180^\circ$

$\therefore \angle CEF = 180^\circ - 90^\circ - 45^\circ = 45^\circ = \angle ECF$

By Triangle with Two Equal Angles is Isosceles, $CF = FE$.

Since $CDEF$ is a rectangle with two adjacent equal sides, it must also be a square.

$\blacksquare$