Tuesday, September 23, 2014

Section 3-6: Constructing Perpendiculars (Day 33)

Section 3-6 of the U of Chicago text deals with constructions. So, we're finally here. The students will need a straightedge and compass to complete this lesson.

Here's a good point to ask ourselves, which constructions do we want to include here? The text itself focuses on the constructions involving perpendicular lines. Well, let's check the Common Core Standards, our ultimate source for what to include:

Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.
Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle.
But let's also go back to what David Joyce writes about constructions:
The book [Prentice-Hall 1998 -- dw] does not properly treat constructions. Constructions can be either postulates or theorems, depending on whether they're assumed or proved. For instance, postulate 1-1 above is actually a construction. On pages 40 through 42 four constructions are given: 1) to cut a line segment equal to a given line segment, 2) to construct an angle equal to a given angle, 3) to construct a perpendicular bisector of a line segment, and 4) to bisect an angle. Later in the book, these constructions are used to prove theorems, yet they are not proved here, nor are they proved later in the book. There is no indication whether they are to be taken as postulates (they should not, since they can be proved), or as theorems. At the very least, it should be stated that they are theorems which will be proved later.
David Joyce, after all, emphasizes that at the very least, constructions should be proved. He writes here that they can be proved later -- but of course, he prefers that theorems not be stated until they can be proved.
So which of the theorems in the Common Core list can be proved so far? Let's look back at that list one by one:
Copying a segment: This should be trivial to construct and prove. The student simply uses the straightedge to draw a line, marks a point O on it, and opens up the compass to the length of the given line segment AB to mark the second point P. The proof that these segments AB and OP have the same length simply follows from the definition of straightedge and compass. It's a bit surprising that the U of Chicago text doesn't begin with this as the first construction, as this one should be easy for the students.
Copying an angle: This is the one that we can't prove yet. The usual construction requires SSS to prove. This is one reason why Joyce would prefer that Chapter 8 of his text occur before these constructions in his Chapter 1.
Bisecting a segment; constructing perpendicular lines, including the perpendicular bisector of a line segment: This is the focus of Section 3-6 of the U of Chicago text. Notice that as soon as we've constructed the perpendicular bisector, we've already done the other two constructions (bisecting the segment and drawing its perpendicular). And so, as soon as we prove the perpendicular bisector construction, we are done.
Given: Circle A contains B, Circle B contains A, Circles A and B intersect at C and D.
Prove: Line CD is the perpendicular bisector of AB
Proof (in paragraph form -- can be converted to two columns later):
Just as in the proof of Euclid's first theorem (Section 4-4, two weeks ago), since both B and C lie on circle A, AB = AC by the definition of circle, and since both A and C lie on circle B, AB = BC. Then by the Transitive Property of Equality, AC = BC -- that is, C is equidistant from A and B. So, by the Converse of the Perpendicular Bisector Theorem, C lies on the perpendicular bisector of AB. In the same way, we can prove that D also lies on the perpendicular bisector of AB. And through the two points C and D there is exactly one line -- and that line is the perpendicular bisector of AB. QED
In many texts, it's pointed out that the compass opening for the two circles need not be exactly the same as AB. All that's necessary is for the opening to be greater than half of AB -- that guarantees that the two circles intersect in two points. 
The text states that the midpoint of AB has been constructed as a "bonus" -- so we've bisected the segment, as requested. All we need now is to construct perpendicular lines -- and that's exactly what the text does in the next example, construct a perpendicular to line AP through point P on the line, using our perpendicular bisector algorithm as a subroutine.
Bisecting an angle: This is an interesting one. As I mentioned last week, Questions 16 and 17 from Section 4-7 of the text show us how to perform an angle "bisectomy." Notice that this construction is slightly different from the one usually given in texts, since the usual construction requires SSS to prove while this one depends only on reflections:
Given: Circle O contains A, Circle O intersects ray OB at C, Line PQ is the perpendicular bisector of AC
Prove: Ray OP is the angle bisector of Angle AOB
Since both A and C lie on circle O, AO = CO by the definition of circle. So, by the Converse of the Perpendicular Bisector Theorem, O lies on the perpendicular bisector of AC. Now let line OP be a reflecting line. The image of A is C, and points O and P, both lying on the mirror, are each their own image. So the image of angle AOP is angle COP. Since reflections preserve angle measure, AOP and COP have the same measure. Therefore, by the definition of angle bisector, ray OP is the angle bisector of AOB. QED
Since we are jumping around the book right now, we can perform a few more constructions that we've skipped over. In Lesson 4-1, we construct the reflection image of point B over the line m:
Given: Circle B intersects m in Q and S, Circles Q and S contain B and B'
Prove: B' is the reflection image of B over m
Once again, using the definition of circle and the Transitive Property of Equality, we have both BQ = B'Q and BS = B'S, so once again, by the Converse of the Perpendicular Bisector Theorem, both Q and S lie on the perpendicular bisector of BB'. Since two points determine a line, the line containing Q and S -- line m -- is the perpendicular bisector of BB'. So by the definition of reflection, B' is the mirror image of B over m. QED
The book states that there is another "bonus" -- line BB' is the perpendicular to m through B. So we have another construction and the existence of a line perpendicular to a line through a point not on the given line. (This line is unique by the Uniqueness of Perpendiculars Theorem.) But Common Core asks for the similar construction:
Constructing a line parallel to a given line through a point not on the line: Believe it or not, we can complete this construction right now. Usually, the construction involves copying in angle in a way so that the angle and its copy are corresponding or alternate interior angles, so that the lines are parallel. We have neither given the angle copy construction nor the parallel test for corresponding angles. But we do have the theorem from yesterday -- the Two Perpendiculars Theorem.
So given a line l and a point P not on the line l, we construct the line parallel to l through point P as follows:
Step 1. Subroutine m, a line perpendicular to l.
Step 2. Subroutine k, a line perpendicular to m through P.
then by the Two Perpendiculars Theorem, k and l are parallel. Notice that line m need not contain point P. If it does, then Step 1 is a "construct perpendicular through point not on line" step and Step 2 is "construct perpendicular through point on line." Or one could choose a point Q on l, and then do the "construct perpendicular through point on line" step first instead of second.
I was not able to create a good worksheet for constructions. So I leave you on your own to find a good worksheet.

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