# What is the quadrilateral formed by the angle bisectors of a parallelogram?

I have drawn a few parallelograms and their angle bisectors in Geometer's Sketchpad. The quadrilateral looks to me to be a rectangle but how can I prove it ?

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Please post a diagram with correct labels. Not sure what you are talking about. –  Will Jagy Mar 31 '13 at 19:14
@WillJagy - added! –  devcoder Mar 31 '13 at 19:18
Thank you. That's a good diagram. –  Will Jagy Mar 31 '13 at 19:24

Let $P$ and $Q$ be adjacent corners of a parallelogram, and let those angles have measure $p$ and $q$. Let $R$ be the point at which the angle bisectors at $P$ and $Q$ meet.

In $\triangle PQR$, we have

$$180^\circ = \angle R + \angle RPQ + \angle RQP = \angle R + \frac{1}{2}p + \frac{1}{2}q = \angle R + \frac{1}{2}\left( p+q \right)$$

Adjacent angles in a parallelogram are supplementary, so $p+q=180^\circ$. Thus,

$$180^\circ = \angle R + 90^\circ \qquad \implies \qquad \angle R = 90^\circ$$

which is to say: Adjacent angle bisectors in a parallelogram meet at right angles.

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@DonAntonio: (a) The diagram was posted while I was composing my answer, so I didn't notice it; and (b) $P$ and $Q$ make a generic pair of adjacent vertices (with generic intersection $R$), so the OP's labeling doesn't matter. –  Blue Mar 31 '13 at 19:23
Good point, @Blue. –  DonAntonio Mar 31 '13 at 19:27

Given two lines intersecting at a point there are two angle bisectors, and these are perpendicular to each other. Up to parallel translation, that's all.

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@hardmath: See my edit. –  Christian Blatter Aug 10 '14 at 15:34