Problem — The bisectors of a rectangle form a square

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Let $EFGH$ be a rectangle (not a square). Prove that the bisectors of its four interior angles determine a square.

(original figure: rectangle $EFGH$ with the four bisectors and the square they determine)

Solution

Since a rectangle is in particular a parallelogram, by the result on the bisectors of a parallelogram the quadrilateral determined by the four bisectors is already a rectangle. Let $P$, $Q$, $R$, $S$ be the intersections of the bisectors relative to the pairs of consecutive vertices $(E,F)$, $(F,G)$, $(G,H)$, $(H,E)$: thus $PQRS$ is a rectangle. It remains to show that two of its consecutive sides are congruent.

Consider the axis of symmetry $\sigma$ of the rectangle perpendicular to the sides $EF$ and $GH$ (the line through the midpoints of $EF$ and of $GH$). The reflection in $\sigma$ swaps $E\leftrightarrow F$ and $H\leftrightarrow G$, and maps the bisector of each angle onto the bisector of the corresponding angle. Consequently:

• $P$, intersection of the bisectors of $E$ and $F$, lies on the axis $\sigma$ and is therefore fixed: $\sigma(P)=P$;
• $Q$ (bisectors of $F$ and $G$) is mapped to $S$ (bisectors of $E$ and $H$), i.e. $\sigma(Q)=S$.

Since a reflection preserves distances and $\sigma(P)=P$, $\sigma(Q)=S$, the segment $PQ$ is mapped to the segment $PS$: hence $PQ\cong PS$. But $PQ$ and $PS$ are two consecutive sides of the rectangle $PQRS$: a rectangle with two consecutive sides congruent is a square. $\;\blacksquare$