Proving that a parallelogram has equal pair of sides?

Answer 1

First draw a parallelogram. I cannot draw one here so I will have to describe the picture and you should draw it. Let ABCD be a parallelogram. I put A on the bottom left, then B on the bottom right, C on the top right and D on the top left. Of course the arguments must apply to an arbitrary parallelogram, but just so you can follow the proof, that is my drawing. Now draw a segment from A to C. It is a diagonal. AB is parallel to CD and AD is parallel to BD because a parallelogram is a quadrilateral with both pair of opposite sides parallel. Now ABC and CDA both form triangles. Let angles 1 and 4 be the angles created by the diagonal and angle BCD of the parallelogram. Angle 1 is above and angle 4 is below. Similarly, let angles 3 and 2 be created by the intersection of the diagonal and angle DAB or the original parallelogram. Now angles 1 and 2 are congruent as are 3 and 4 because if two parallel lines are cut by a transversal, the alternate interior angles are congruent. Next using the reflexive property AC is congruent to itself. Now triangle ABC is congruent to triangle CDA by Angle Side Angle (SAS). This means that AB is congruent to CD and BC is congruent to AD by corresponding parts of congruent triangles are congruent (CPCTC). So we are done!