Problem — The bisectors of a parallelogram form a rectangle

Text

Let $WXYZ$ be a parallelogram. Prove that the bisectors of its four interior angles, by intersecting, determine a rectangle.

(original figure: parallelogram $WXYZ$ with the four bisectors and the rectangle they determine)

Solution

In a parallelogram two consecutive angles are supplementary; in particular $\widehat{W}+\widehat{X}=180^\circ$. Let $w,x,y,z$ be the bisectors of the angles at $W,X,Y,Z$. The bisectors $w$ and $x$, relative to the consecutive vertices $W$ and $X$, meet at a point $P$. In triangle $WXP$: $\widehat{PWX}=\tfrac12\widehat{W},\qquad \widehat{PXW}=\tfrac12\widehat{X},$ hence $\widehat{PWX}+\widehat{PXW}=\tfrac12(\widehat{W}+\widehat{X})=\tfrac12\cdot 180^\circ=90^\circ,$ so that $\widehat{WPX}=180^\circ-90^\circ=90^\circ$. By the same reasoning the bisectors relative to the other pairs of consecutive vertices meet at $Q$ (from $X$ and $Y$), $R$ (from $Y$ and $Z$), $S$ (from $Z$ and $W$), forming a right angle at each point. In the quadrilateral $PQRS$ the angle at $P$ is $\widehat{SPQ}$, vertical to $\widehat{WPX}$, hence also a right angle; in the same way the angles at $Q$, $R$, $S$ are right. A quadrilateral with four right angles is a rectangle. $\;\blacksquare$