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Congruent triangles:

Take a parallelogram PQRS.

Draw in the diagonals PR and QS.

Let the point where the diagonals meet be M,

Consider one pair of the parallel sides, PS and QR, say.

Consider angles PSQ and RQS:

As PS and QR are parallel they are equal (Z- or alternate angles).

Now consider angles SPR and QRP:

As PS and QR are parallel they are equal (z- or alternate angles).

As PS and QR are opposite sides of a parallelogram they are equal in length; thus the triangles PMS and RMQ are congruent (Angle-Angle-Side).

As the two triangle are congruent, equivalent sides are equal in length.

Thus QM is the same length as MS and PM is the same length as MR

As QM is the same length as MS and QMS lie on a straight line, M must be the mid point of QS, ie the diagonal PQ bisects the diagonal QS

Similarly PM is the same length as MR and PMR lie on a straight line, thus M must be the mid point of PR, ie the diagonal QS bisects the diagonal PR

Therefore the diagonals of a parallelogram bisect each other.

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Q: Which concept can be used to prove that the diagonals of a parallelogram bisect each other?
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