**Theorem 1:** If the diagonals of a quadrilateral bisect each other then the quadrilateral is a parallelogram.

**Proof: **Let the quadrilateral be *ABCD* with diagonals *AC* and *BD* intersecting at *P*:

Then *AP* = *PC* and *DP* = *PB* since the diagonals bisect each other. Now consider triangles *APD* and *CPB*. The vertical angles *APD* and *CPB* are equal, and we have pairs of sides that are equal (*AP* = *PC* and *DP* = *PB*), so these triangles are congruent by **SAS**. As a consequence, the corresponding angles, *DAP* and *BCP* are equal. But these are alternate interior angles for lines *AD* and *BC* with transversal *AC*. So this proves side*AD* is parallel to side *BC*.

Similarly, if we consider triangles *APB* and *CPD*, they are congruent by the same reasoning, so corresponding angles *BAP* and *DCP* are equal. These are alternate interior angles for lines *AB* and *DC*, so those lines are also parallel.

Therefore *ABCD* is a parallelogram, by definition.