Inscribed Squares in Right-Angled Triangle/Compass and Straightedge Construction/Side Lies on Hypotenuse

Theorem

 * Inscribed_square_hc.png

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

Construct a perpedicular line from $C$ to side $AB$, and name the intersection $D$.

Construct the angle bisector of $\angle ADC$.

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

Construct a line parallel to $AB$ that passes through $E$ to meet $CD$ at $F$.

Extend $AF$ to side $BC$ at $G$.

Construct a line parallel to $AB$ that passes through $G$ to meet side $AC$ at $H$.

Construct perpedicular lines from $H$ and $G$ to side $AB$, and name the intersections $I$ and $J$ respectively.

Then $GHIJ$ is an inscribed square.

Proof
Note that $HG \perp HI$, $HG \perp GJ$ and $HG \parallel IJ$ by construction.

Therefore $GHIJ$ is a rectangle.

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

Since $\angle EFD = 90^\circ$ by construction, by Sum of Angles of Triangle equals Two Right Angles:
 * $\angle DEF + \angle EFD + \angle FDE = 180^\circ$


 * $\therefore \angle DEF = 180^\circ - 90^\circ - 45^\circ = 45^\circ = \angle FDE$

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

Since $EF \parallel HG$, $\triangle AEF \sim \triangle AHG$ by Equiangular Triangles are Similar.

Similarly, since $FD \parallel GJ$, $\triangle AFD \sim \triangle AGJ$ by Equiangular Triangles are Similar.

By definition of similar triangles:
 * $\dfrac {EF} {HG} = \dfrac {AF} {AG} = \dfrac {FD} {GJ}$

As $EF = FD$, we must have $HG = GJ$.

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