Lesson 7-7: Sufficient Conditions for Parallelograms (Day 69)

Today I subbed in a seventh-grade math class. The class was solving simple one-step equations. Of course, the most common errors were sign errors -- this is not surprising, especially since seventh graders are seeing negative numbers for the first time. It didn't help that many of the questions on the worksheet were written awkwardly, such as y - -20 = 7 (with no parentheses). Fortunately, the students had already learned that "minus minus" becomes "plus plus," but this sort of equation is still very confusing at this level.


This is what I wrote last year about today's Geometry lesson:

Lesson 7-7 of the U of Chicago text is on sufficient conditions for parallelograms. This is how we prove that a figure is a parallelogram. Dr. Franklin Mason would call these the "parallelogram tests," or, as he often abbreviates it, the "pgram tests."

The sufficient conditions for a parallelogram are that if a quadrilateral has opposite sides congruent, or if it has opposite angles congruent, or if its diagonals bisect each other, or if it has one pair of sides both congruent and parallel, then the figure is a parallelogram.

Think about it. Suppose we are given a quadrilateral with opposite sides congruent. It's easy to use SSS to prove that it's a parallelogram. The two pairs of given sides already give us SS, and for the third S, we draw in a diagonal and note that it's congruent to itself. And so we use the traditional SSS proof to prove this theorem.

In general, if we are given that a figure is symmetrical -- because it's an isosceles triangle or a kite or any figure that has its own Symmetry Theorem -- then we just use that theorem. But if we're given things like congruent sides or angles, then these are probably the S or A parts of one of the traditional Congruence Theorems SSS, SAS, or ASA and so we should use those.

Therefore this section is an excellent demonstration of the power of the SSS, SAS, and ASA Congruence Theorems, even in a Common Core class where the focus is on transformations.

Parallelograms don't have reflection symmetry. On the other hand, they do have rotational symmetry, and here is an example of a proof where rotational symmetry comes in handy:

-- Prove that if two perpendicular lines intersect at the center of a square, then they divide that square into four congruent regions.

Let's call the square ABCD. A few cases are easy to prove. Suppose the two perpendicular lines are the diagonals of the square AC and BD, which meet at the center O. These lines are perpendicular because the diagonals of any kite are perpendicular, and a square is a kite. These lines intersect at the center O because the diagonals of any rectangle are congruent and bisect each other, and a square is, of course, a rectangle. Then this is enough to prove that the triangles AOB, BOC, COD, and DOA are congruent by SAS.

Another easy case is if the two lines are the perpendicular bisectors of the four sides. Each of the four regions has three right angles (one an angle of the square, and two more right angles by the definition of perpendicular) and hence four right angles (since the sum of the angles of a quadrilateral is 360), and hence each is a rectangle. And as each one's length and height is half that of the original square (by the definition of bisector), hence each is a square. So there are four congruent squares, each with a side length half that of the original square.

But the tough case is when the two lines are neither diagonals nor perpendicular bisectors. The four points where the two lines intersect AB, BC, CD, and DA respectively, let's call these points E, F, G, and H respectively. These quadrilaterals -- AEOH, BFOE. CGOF, and DHOG -- are not special quadrilaterals like kites or trapezoids, yet we must prove them congruent.

Here's my claim -- a 90-degree rotation about O maps AEOH to BFOE. The proof -- we should have proved earlier this week (our extra discussion for Section 7-6) that a rotation maps regular polygons to themselves, so ABCD is mapped to BCDA. Clearly O, the center of rotation, is mapped to itself, so it remains to show where points such as E are mapped. Since EG and FH are perpendicular, rotating line EG must give us line FH, so E'is clearly on line FH. Since E is on AB and AB maps to BC, E' must lie somewhere on line BC. And BC and FH intersect at F, so E' is F.

Similarly, F' is G, G' is H, and H' is E.Therefore the rotation maps AEOH to BFOE -- and since there is an isometry mapping AEOH to BFOE, they are congruent. Similarly both are congruent to CGOF and DHOG. QED

This question was inspired by one I saw many years on the Theoni Pappas calendar. There's another interesting congruence question that was inspired by one I saw on the test of a geometry student I was tutoring a few years ago, and I'll give that question next time.