Friday, December 11, 2015

Activity and Review: The Triangle Exterior Angle Equality (Day 70)

Today I subbed in another math class, this time for eighth graders. (Yes, I know -- no math for a month, and then thrice in a week!) The students were working on a graphing activity -- the kind of activity that teachers often assign just before Thanksgiving or winter break (even though the holidays are still over a week away). It is a simple graphing activity, not one with dilations like the one I posted before Thanksgiving. I will post this activity for you on the blog on the actual last day before winter break. As it turns out, this worksheet contains points in all four quadrants, but no negative numbers.

Last year at this time, I posted a semi-activity, semi-review on the Triangle Exterior Angle Equality. I point out that in some ways, this lesson still makes sense at this point. In the next few weeks, we'll be covering the last part of Chapter 13, which is where this theorem appears in the U of Chicago text. In that text, it is the gateway to the closely related Triangle Exterior Angle Inequality, or TEAI, which is used to prove various inequalities in triangles.

This is what I wrote last year about this worksheet:

I decided to include questions on exterior angles, congruent triangles, as well as explanations for why two triangles are or aren't congruent

Meanwhile, here's the congruence question that I promised you from yesterday. As I mentioned, this is based on a question asked on a test for one of the students I tutor:

What type of quadrilateral has diagonals dividing it into four congruent triangles?

The correct answer is a rhombus, but how do we prove this? Again, we may refer to the quadrilateral as ABCD and the point where AC and BD intersect as O. But the problem is that we only know that AOB is congruent to the triangle with vertices BO, and C -- but is it triangle BOC, or is it COB, or possibly even BCO? Nowhere in the question is it stated which vertices correspond -- which nearly every congruence problem states!

We ask, to which angle in triangle BOC does angle AOB correspond? We notice that AOB is actually an exterior angle of triangle BOC -- which means that it's greater than the measure of either of its remote interior angles, BCO and CBO. (Hey, there's that Triangle Exterior Angle Inequality again, but notice that only the Inequality is needed for this one, not the full Equality.) So AOB can't be congruent to either of them, hence it can't correspond to either of them. So AOB must correspond to angle COB, and since these congruent angles form a linear pair, they must be right angles.

Now we know that AOBBOCCOD, and DOA are congruent right triangles, but notice that we still don't know which sides correspond. But notice that in triangles AOB, we don't know whether leg AO corresponds to BO or CO of the next triangle. Yet we do know that the sides opposite the known right angle -- that is, the hypotenuses -- must correspond! So AB is congruent to BC, and likewise to CD and DA, the other hypotenuses. These four hypotenuses are the four sides of the quadrilateral ABCD, and they are all congruent. Therefore, the figure is a rhombus. QED

Technically speaking, we should prove the converse as well. If a quadrilateral is a rhombus, then its diagonals divide it into four congruent triangles. But this follows easily from well-known properties of the rhombus.

This whole proof is somewhat advanced for a geometry student -- so I don't know exactly what the teacher was expecting the students to do. (Just guess it's a rhombus and prove the converse?) The student ended up writing "square" and didn't write much of a proof -- yet the teacher must have been feeling super-nice and awarded him 3 out of 4 points! (He's clearly not a Putnam grader!)

Returning to the present, I decided to include only the first page from last year. There's one more topic that appears in the Common Core yet is not covered well in the U of Chicago text:

Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle.

There are many constructions throughout the U of Chicago text, but not any of these. Yet in this week's Lesson 7-6, it's the Center of a Regular Polygon Theorem that tells us that there even exists a circle that passes through the vertices of any regular polygon. There are even pictures of the first two in the text -- an equilateral triangle and a square inscribed in circles -- yet the text doesn't direct students to construct them.

Last year, I called the worksheet I posted an "activity," yet it's not much of an activity -- I called it that because it referred to a day when I was subbing in the classroom and I turned it into a game by playing my usual "What's my age?" game. But now that I'm adding constructions to it, it's more of an actual activity since students are drawing with straightedge and compass. We can even make it into a construction game -- yes, that would be Euclid: The Game.

The picture of the inscribed square in the U of Chicago text gives away its construction. We begin by drawing a circle and a diameter, then we find the perpendicular bisector of this diameter, which contains another diameter of the circle. The two diameters are the diagonals of the square.

The regular hexagon is actually the easiest to construct -- this is because the side length of the hexagon equals the radius of the circle. So we start at any point on the circle and open up the compass to the radius of the circle, then use that setting to construct five more points on the circle. Then the equilateral triangle is formed by skipping every other vertex of that hexagon.

No comments:

Post a Comment