Given that a parallelogram is a quadrilateral with two pairs of parallel sides, prove that a quadrilateral has opposite sides of equal length if and only if it is a parallelogram.
The first part of the proof is to show that a parallelogram has opposite sides equal length.
As AC is parallel with BD we know that the alternate angles, ACB and DBC, are equal. Similarly angles ABC and DCB are equal.
In triangles ABC and BCD they share the diagonal BC and by the angle/side/angle property we determine that they are congruent. Hence length AB = CD and AC = BD.
Now we shall prove the converse, that a quadrilateral with opposite sides equal length must be a parallelogram.
We begin with the line segments AB and AC, and without loss of generality let us suppose that AB AC. Construct two circles: centre B, radius AC; and centre C, radius AB. The intersection of these two circles generates two points, D and E.
It can be seen that these points of intersection, D and E, lie either side of AC. So it is only by connecting B (and C) to D will we produce a simple polygon (connecting B to E will cross the segment AC producing a complex polygon).
By the method of construction we know that AB = CD and AC = BD. Therefore the triangles ABC and BCD have three common lengths and must be congruent.
Because angles ACB and DBC are equal, segments AC and BD must be parallel. Similarly AB is parallel with CD and we prove that ABCD is a parallelogram.
Hence we have proved that a quadrilateral has opposite sides of equal length if and only if it is a parallelogram.