## Tuesday, May 19, 2015

### PARCC Practice Test Question 26 (Day 171)

Question 26 of the PARCC Practice Test is on properties of parallelograms. It requires students to know two different types of proofs: two-column and coordinate:

One method that can be used to prove that the diagonals of a parallelogram bisect each other is shown in the given partial proof:

(The figure shows that PR and QS intersect at T.)

Given: Quadrilateral PQRS is a parallelogram
Prove: PT = RT, ST = QT

Proof:
Statements                      Reasons
1. Quad PQRS is pgram   1. Given
2. PQ | | SR, PS | | QR     2. Definition of parallelogram
3. Angle PQS cong. RSQ, 3. __________
Angle QPR cong. SRQ
4. __________                 4. Opposite sides of a parallelogram are congruent
5. Tri. SRT cong. QPT     5. __________
6. PT cong. RT,               6. Corresponding parts of congruent triangles are congruent
ST cong. QT
7. PT = RT, ST = QT       7. Definition of congruent line segments

Part A

Which reason justifies the statement for step 3 in the proof?

(A) When two parallel lines are intersected by a transversal, same side interior angles are congruent.
(B) When two parallel lines are intersected by a transversal, alternate interior angles are congruent.
(C) When two parallel lines are intersected by a transversal, same side interior angles are supplementary.
(D) When two parallel lines are intersected by a transversal, alternate interior angles are supplementary.

Part B

Which statement is justified by the reason for step 4 in the proof?

(A) PQ is congruent to RS
(B) PQ is congruent to SP
(C) PT is congruent to TR
(D) SQ is congruent to PR

Part C

Which reason justifies the statement for step 5 in the proof?

(A) side-side-side triangle congruence
(B) side-angle-side triangle congruence
(C) angle-side-angle triangle congruence
(D) angle-angle-side triangle congruence

Part D

Another method of proving diagonals of a parallelogram bisect each other uses a coordinate grid.

(The figure shows the following coordinates: P(r, t), Q(r + s, t), R(s, 0), S(0, 0).)

What could be shown about the diagonals of parallelogram PQRS to complete the proof?

(A) PR and SQ have the same length.
(B) PR is a perpendicular bisector of SQ.
(C) PR and SQ have the same midpoint.
(D) Angles formed by the intersection of PR and SQ each measure 90 degrees.

For this question, the first three parts fill in steps of the two-column proof. Part A is clearly about congruent angles and same side interior angles aren't congruent, so the answer is (B). Part B explicitly mentions opposite sides, and the only choice that gives opposite sides is (A). For Part C, it may help to draw in the known pairs of congruent parts, but this will be tricky on the computer. We see that the known sides from Part B are included between the angles, so the answer is (C).

Finally, Part D requires us to look at the coordinate proof. We want to prove that the diagonals of the parallelogram bisect each other -- and once again, we must ask what it means to bisect. That word means to cut in half. So the diagonals must cut each other in half -- at their midpoints. So the answer is in fact choice (C).

For students, proofs are often tricky. Fortunately for them, most of the steps of the proofs are already given and all they have to do is fill in the blanks. One thing that might throw students off is that we're so used to seeing the abbreviations CPCTC, ASA, etc., that students might be thrown off by seeing the whole phrase "Corresponding parts of congruent triangles are congruent" written out.

The main section of the U of Chicago text is Section 7-6, "Properties of Special Figures." This section contains a Properties of a Parallelogram Theorem which states that in any parallelogram, each diagonal forms two congruent triangles, opposite sides are congruent, and that the diagonals intersect at their midpoints. Notice that the U of Chicago doesn't use the phrase "diagonals bisect each other" but instead, it uses the phrase "diagonals intersect at their midpoints," which would make the answer to Part D more obvious.

Speaking of Part D, of course a coordinate proof of this theorem couldn't appear in the U of Chicago until Chapter 11 on coordinate geometry. This specific proof doesn't appear in the text -- I suppose if it were to appear, it would be in Section 11-4, since this is the section on midpoints,

I've said many things about proofs on this blog, and I've quoted many authors repeatedly -- Drs. David Joyce, Franklin Mason, and Hung-Hsi Wu. Each of them has his own philosophy of what proofs should look like. But this is a Common Core blog, and my job is to make sure that students are prepared for the Common Core tests. Therefore, what actually appears on the PARCC and SBAC exams have priority over all other considerations. It doesn't matter how persuasive or elegant are the arguments of these authors, or those of the U of Chicago or any other text, if what they say doesn't correspond to what actually appears on the PARCC or SBAC exams.

Joyce writes that anything that hasn't been proved yet should be avoided until it can be proved -- especially coordinates. Yet here we have a coordinate proof right on the practice PARCC. Wu tries to create a proof that coordinates actually work. In order to get coordinates to work, Wu first proves his Fundamental Theorem of Similarity (his Theorem 22) -- in particular, he must prove the special case (Lemma 13) when the scale factor is an integer. But his proof of Lemma 13 uses a previously proved result, Theorem 16 -- which just happens to be the statements that the diagonals of a quadrilateral bisect each other if and only if it is a parallelogram!

So we see obviously circularity here. In order to give a coordinate proof of Part D, we must first prove that coordinates work. To prove that coordinates work, we must prove Wu's Lemma 13. To prove Lemma 13, we must prove Wu's Theorem 16. But Theorem 16 is exactly what we're trying to prove in Part D!

Of course, the situation isn't completely circular. If we really need to prove Theorem 16, we could just use the two-column proof given in Parts A-C. Then again, the proof of Theorem 16 isn't actually the Properties of a Parallelogram Theorem that we're proving in this problem, but rather its converse, which is part of the "Sufficient Conditions for a Parallelogram Theorem" of Section 7-7 (and the "pgram tests" of Dr. M). But Theorem 16 is stated as a biconditional.

If we think about it, we shouldn't be surprised that Wu's Theorem 22 is so difficult to prove. We are trying to show that the dilation image of AB has length r * AB, where r is the scale factor. But until we reach the section on dilations, all of the theorems are about congruence, so how can we use congruence to prove that something has length r * AB? If r is an integer, then we can divide A'B' into r congruent segments of length AB -- and this is essentially what Wu does. If r = 1/n for some natural number n, then we can divide the original segment AB into n congruent segments of length A'B'. But if r is irrational, then we really can't show that A'B' has length r * AB using only congruence.

So Wu only really proves his theorem for rational scale factors r. He justifies this by introducing a "Fundamental Assumption of School Mathematics," which basically states that anything high school students prove true for all rational numbers is true for all real numbers. In other words, after spending all that effort trying to prove Theorem 22 for just certain (i.e., rational) values of r, he simply makes an assumption -- and in Geometry, we call assumptions postulates -- and declares the Theorem to be proved for all values of r.

But then we wonder -- why can't we just declare Wu's Fundamental Theorem of Similarity to be a postulate and then dispense with his long proof altogether? We've mentioned earlier that if a theorem is too hard for students to prove, we should just declare it to be a postulate. And besides -- we already have a Reflection Postulate (Chapter 4 in the U of Chicago). We don't need a Translation or Rotation Postulate because translations and rotations are composites of reflections, so they are already covered by the Reflection Postulate. But dilations aren't composites of reflections, so we could have a separate postulate, a Dilation Postulate, to describe their properties.

Mathematicians avoid declaring new postulates out of statements that can be proved. But even Joyce himself writes about similarity:

Appropriately for this level, the difficulties of proportions are buried in the implicit assumptions of real numbers.

So even he would allow Geometry students to make assumptions (postulates) about real numbers. The proof of Wu's Theorem 22 can be proved for all rational numbers, but not for all real numbers. And besides, is Wu's proof of Theorem 22 really appropriate for high school students? That proof is very difficult and would end up only discouraging students from studying further proofs, including the proofs that actually matter (such as the proofs in Parts A-D of today's featured PARCC question).

So it's settled! We need a Dilation Postulate to simplify the proofs about dilations. Now what should our Dilation Postulate. On one hand, our Reflection Postulate contained four parts -- and these are already labeled A-B-C-D (Angle measure, Betweenness, Collinearity, Distance). But on the other hand, mathematicians prefer assuming less and proving more. The one property that we surely need to postulate is Distance, as this is the part involving real numbers. If we look at Section 12-3 of the U of Chicago text, where the properties of dilations are discussed, we see the following theorem:

Size Change Distance Theorem:
Under a size change with magnitude k > 0, the distance between any two image points is k times the distance between their preimages.

Yes, I know that I just switched from Wu's variable r to the U of Chicago's k. Anyway, the text states that a proof of this theorem was given in Section 12-1, but that's the coordinate proof that we consider to be circular since we need dilations before we can have coordinates. And so this is the statement that we will make into a postulate. We want to use the term "dilation" rather than "size change" and this is now a postulate, not a theorem. So the new name of this statement is "Dilation Distance Postulate":

Dilation Distance Postulate:
Under a dilation with scale factor k > 0, the distance between any two image points is k times the distance between their preimages.

Notice that if one of the preimage points is the center of the dilation, then this statement follows directly from the definition of dilation. It's only when neither point is the center when the Dilation Distance Postulate is necessary.

But what about Angle measure, Betweenness, and Collinearity? We notice that the next theorem in the U of Chicago text states that dilations preserve Betweenness and Collinearity. This proof is not difficult at all, as it follows right from what we are now calling the Dilation Distance Postulate. So we don't need to include Betweenness and Collinearity in our Postulate, as we can just prove them using the U of Chicago proof.

Before we can get to Angle measure, we must consider parallel lines -- and parallel lines and congruent angles are closely related. The statement that a line and its dilation image are parallel is important, as this is explicitly mentioned in the Common Core Standards and we've already seen a PARCC question about this (Question 13). But the theorem that establishes this in the text is said to have been proved in Section 12-1 -- meaning the circular coordinate proof. Does this mean that we should assume this statement in our Postulate?

Back when we covered PARCC Question 13, we discussed an actual proof of this statement. It was based on a proof in Wu about 180-degree rotations that also works for dilations. We first begin by establishing that any line passing through the center is invariant. Then we can prove using an indirect proof that the image of any line not containing the center is parallel to its preimage. This is what I wrote earlier:

Assume that the line and its image are not parallel -- that is, that they have a point in common. That is, there exists a point A on the line such that its image A' is also on the line (that's where the line and its image intersect). So the original line can now be written as line AA'. Notice that O is also on line AA' -- once again, it follows from first case above that the points OA, and A' are collinear. But this contradicts the assumption that O is not on the original line. Since the assumption that the line and its image intersect leads to a contradiction, we conclude that the line and its image are parallel.

As long as students can do indirect proofs, this can be taught in the dilation lesson. If we wish to avoid indirect proofs, then we would have to include the statement that a line and its dilation image are parallel as part of the Dilation Postulate. Finally, the statement that dilations preserve Angle measures
is proved in the U of Chicago text using the statement about parallel images, so we can keep that one.

At this point, we should be able to use dilations to establish similarity, use similarity to prove the Pythagorean Theorem (distance formula) and slope formulas, use the distance formula to prove the midpoint formula, and finally use the midpoint formula to complete the proof in Part D. Nothing is circular, and everything theorem follows either from a previously proved theorem or ultimately back to the Dilation Distance Postulate.

There are a few points left to settle here. First of all, we've only used 2D dilations, but if we want to discuss 3D figures that are similar (in order to prove that one figure has k^3 times the volume, but only k^2 times the surface area of the other), then we need 3D dilations. The only problem with 3D dilations is that our proof that a line and its image are parallel worked by assuming that they intersect and deriving a contradiction. But in 3D there's a third possibility -- skew lines. As the proof is written, there's nothing preventing a line and its dilation image from being skew lines.

In Section 9-1, we have two new parts of the Point-Line-Plane Postulate:

e. If two points lie in a plane, the line containing them lies in the plane.
f. Through three noncollinear points, there is exactly one plane.

These guarantee that a line and its dilation image can never be skew lines. If O is the center of dilation and AB is the preimage, part (f) guarantees exactly one plane containing O, A, and B. (We don't have to worry about O, A, and B being collinear because if they were, by the definition of dilation, A' and B' would lie on that same line.) By the definition of dilation, A' lies on line OA, and so by part (e), all of line OA, including A', lies in plane OAB. Likewise B' also lies in plane OAB. Finally, since A' and B' both lie in this plane, applying (e) again, all of line A'B' lies in plane OAB. Thus there is a plane, namely OAB, containing AB and A'B'. Therefore AB and A'B' are coplanar, not skew.

By this point, you might be wondering, how did we get from a simple proof about parallelograms to a long discussion about dilations? The point is that Part D was a coordinate proof, and as soon as we have coordinates, we have dilations and similarity. Let's review the specific Common Core Standards which require us to use similarity to derive the Pythagorean Theorem (and thus the Distance Formula) and other properties of the coordinate plane:

CCSS.MATH.CONTENT.HSG.SRT.B.4
Prove theorems about triangles. Theorems include: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity.

CCSS.MATH.CONTENT.8.EE.B.6
Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b.

And so as soon as I saw a coordinate proof on the test, I had to double check to make sure that all of my previous proofs of the above are solid. I believe that our Dilation Distance Postulate is the foundation on which this sequence of proofs lies.

But there is still one more problem, not just with today's question, but with all of the coordinate problems that have appeared on the PARCC. I say that we must establish transformations -- most notably dilations -- before we can have the coordinate plane. But the easiest way for students to learn and understand transformations is to perform them on the coordinate plane -- especially translations (while only certain reflections, rotations, and dilations can be performed easily on the xy-plane).

Once again, there are only two possibilities. The first is to teach transformations using a coordinate plane and declare by fiat that the transformations actually work, just so that the students can understand transformations. Then later on, we go back and prove that the transformations work. This violates the Joyce ideal that nothing is taught before it can be proved.

The other is to teach transformations without coordinates (as the U of Chicago does in Chapter 6), use them to derive the coordinate plane, and then go back and perform the simplest transformations (translations, reflections in the axes, 180-degree rotations/dilations centered at the origin) on the coordinate plane (which the U of Chicago doesn't do fully, yet is crucial for PARCC). This sequence is logically correct, but it won't work if students have trouble understanding transformations without the benefit of a coordinate plane.

To wrap up today's post, notice that Wu, in giving his version of the proof for Parts A-C, does so simply by rotating the parallelogram 180 degrees. The U of Chicago doesn't give this proof, and neither does the PARCC. It may be helpful for PARCC to know that a parallelogram has 180-degree rotational symmetry, but as far as completing proofs is concerned, it's better to give a more standard proof using triangle congruence.

PARCC Practice EOY Exam Question 26
U of Chicago Correspondence: Section 7-6, Properties of Special Figures
Key Theorem: Properties of a Parallelogram Theorem ("pgram consequences")

In any parallelogram:
c. the diagonals intersect at their midpoints

Common Core Standard:
CCSS.MATH.CONTENT.HSG.CO.C.11
Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals.

Commentary: The U of Chicago gives a two-column proof of this. It differs slightly from the PARCC proof because the U of Chicago is proving all three parts a, b, and c in the same proof, while the PARCC proof assumes part b as one of its reasons. Question 2c is the only question in the exercises specifically mentioning this theorem. In the SPUR section, Question 26 under Properties refers to this theorem. Question 38 is an interesting Uses question that refers to this property, the converse of this theorem, as well.