Problem 1 was given as an in-class group activity. The time allotted as 25 minutes. This theorem is an if-and-only-if, so there are two parts to the solution.

Problem 2 was demonstrated quickly on the overhead and was not done as a group activity.

State the definition of a parallelogram (the one in B&B). Prove that a quadrilateral is a parallelogram if and only if the diagonals bisect each other. (In other words, the diagonals intersect at a point M, which is the midpoint of each diagonal.)

Definition. A quadrilateral ABCD is a parallelogram if AB is parallel to CD and BC is parallel to DA.

Assertion 1. If ABCD is a parallelogram, then the diagonals of ABCD bisect each other.

Let O be the intersection of the diagonals AC and BD. The Assertion can be restated thus: O is the midpoint of AC and also the midpoint of BD.

Since O is on segment AC, O is the midpoint of AC if AO = CO. Likewise, O is the midpoint of BD if BO = DO. This is what we will prove using congruent triangles.

First we show triangle ABO is similar to triangle CDO using Angle-Angle. Since line AC is a transversal of the parallel lines AB and CD, then angle OAB = angle CAB = angle ACD = angle OCD. Also, by vertical angles, angle AOB = angle COD. Thus triangle ABO is similar to triangle CDO.

Next we show that these two triangles are congruent by showing the ratio of similitude is 1. We know from the homework (*) that opposite sides of ABCD, AB = CD. These are two corresponding sides of the similar triangles, so the two triangles ABO and CDO are congruent.

From the congruence, we conclude that AO = CO and BO = DO.

Assertion 2. If ABCD is a quadrilateral such that the diagonals AC and BD bisect each other, then ABCD is a parallelogram.

Let M be the intersection of AC and BD. We are given than M is the midpoint of AC and also of BD, so MA = MC and MB = MD.

We also know that angle AMB = angle CMD by vertical angles. Thus we conclude that triangle AMB is congruent to triangle CMD by SAS.

Corresponding angles are congruent. Thus angle MAB (which is the same as angle CAB) and angle MCD (which is the same as angle ACD) are congruent. Since AC is a transversal of lines AB and CD, these equal alternate interior angles imply that the lines AB and CD are parallel.

Thus we see that two opposite sides of ABCD are parallel. Since there was nothing special about those two side, using the same argument, we can also conclude that BC and DA are parallel, so by definition ABCD is a parallelogram.

Proof: In the homework, it was proved that if a quadrilateral ABCD has opposite sides equal, then it is a parallelogram. This follows from that result.

(b) Prove that a parallelogram with perpendicular diagonals is a rhombus.

Proof: From Problem 1, we know that the diagonals of a parallelogram ABCD bisect each other. Let M be the intersection of the diagonals. We know from this that MA = MC and MB = MD.

If we also assume that AC is perpendicular to BC, then each of the angles AMB, AMD, CMB, and CMD are right angles. Thus the triangles AMB, AMD, CMB, and CMD are congruent by SAS. From this is follows that the hypotenuses are all congruent: AB = AD = CB = CD. This says ABCD is a rhombus, by definition.

Proposition: If ABCD is a parallelogram, its opposite sides are equal.

Proof. By definition, line AB is parallel to line CD and line BC is parallel to line DA. We must prove that AB = CD and BC = DA.

We will prove that triangle ABC is congruent to triangle CDA by ASA. The two triangles have a common side AC = CA.

To prove the angles congruent, we use transversals. The line AC is a transversal of parallel lines AB and CD, so angle CAB is congruent to angle ACD. Also line AC is a transversal of parallel lines BC and DA, so angle ACB is congruent to angle CAD.

Thus by ASA, triangles ABC and CDA are congruent. Corresponding sides are equal, so AB = CD and BC = DA.

[Extra credit opportunity. The first person to email to the Math 444-487 email to say what words the initials Q.E.D stand for and what they mean gets extra credit.]