91Ó°ÊÓ

Draw a parallelogram ABCD with diagonals AC and BD, intersecting at point E.

In parallelogram ABCD, we have the following properties: 1. Opposite sides are parallel: AB is parallel to CD, and AD is parallel to BC. 2. Opposite sides are congruent: AB = CD and AD = BC. 3. Opposite angles are congruent: angle A is congruent to angle C, and angle B is congruent to angle D.

We will now show that triangle ABE is congruent to triangle CDE, and triangle ADE is congruent to triangle BCE. Using the properties listed above, we can see that triangle ABE and triangle CDE have: 1. Side AD = BC (opposite sides of a parallelogram are congruent) 2. Angle ABE is congruent to angle CDE (opposite angles of a parallelogram are congruent) 3. Side AB = DC (opposite sides of a parallelogram are congruent) Since we have two congruent sides and the included angle, we can use the Side-Angle-Side (SAS) postulate to show that triangle ABE is congruent to triangle CDE. Therefore, BE = DE. Similarly, triangle ADE and triangle BCE have: 1. Side BC = AD (opposite sides of a parallelogram are congruent) 2. Angle DAE is congruent to angle BCE (opposite angles of a parallelogram are congruent) 3. Side CD = BA (opposite sides of a parallelogram are congruent) Using the SAS postulate again, triangle ADE is congruent to triangle BCE. Therefore, AE = EC.

Since BE = DE and AE = EC, we can conclude that the diagonals of a parallelogram bisect each other.