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If PQRS is a rectangle, then the angle at P is a right angle, and so the opposite angle at P is also a right angle, as shown in the diagram below. Since the lines drawn are angle bisectors, the green angles are equal, the orange angles are equal and the blue angles are equal.

 
By triangle ABP, the green angle and the orange angle must add up to 90$^\text{o}$.

Similarly, from triangle AQD, which has a right angle at Q, the blue angle and the orange angle must add up to 90$^\text{o}$.

So the green and the blue angles must be the same. Continuing, the angles at A and C must also be equal, so opposite angles of ABCD are equal. Which means that ABCD must be a parallelogram.
You can find more short problems, arranged by curriculum topic, in our short problems collection.

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The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.

NRICH is part of the family of activities in the Millennium Mathematics Project.

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