But there's some big news in the Common Core world. Yesterday, the PARCC finally released hundreds of test questions, including many in Geometry. So it goes without saying that I will devote most of this post to discussion of the Released Test Questions:

https://prc.parcconline.org/system/files/Geometry%20-%20PBA%20-%20Item%20Set_0.pdf

https://prc.parcconline.org/system/files/Geometry%20-%20PBA%20-%20Key.pdf

https://prc.parcconline.org/system/files/Geometry%20-%20EOY%20-%20Item%20Set.pdf

https://prc.parcconline.org/system/files/Geometry%20-%20EOY%20-%20Key.pdf

There are a total of 52 Geometry questions available -- 18 for the Performance-Based Assessment (PBA), and 34 for the End-of-Year Assessment (EOY). Recall that the PARCC is considering combining the PBA and EOY into a single test, so this year's students may only have to take one of the tests this year, not both.

Just like last year, my focus will be on the EOY. But I'll start with the PBA. For each of the 18 items on the PBA, I will give the topic and the corresponding chapter of the U of Chicago text in which the topic is (or ought to be) covered:

1. Dilations, Chapter 12

2. Square Diagonals/Trig, Chapter 14

3. Definition of Circle, Chapter 2

4. Dilations, Chapter 12

5. Dilations, Chapter 12

6. Definition of Tangent/Trig, Chapter 14

7. Similarity, Chapter 12

8. Angle of Elevation/Trig, Chapter 14

9. Dividing Into Parts (1D number line), Chapter 1?

10. Definition of Segment, Chapter 1

11. Area of Rectangle/Algebra, Chapter 8

12. Circumference of Circle/Rates, Chapter 8

13. Hexagon Rotational Symmetry, Chapter 5

14. Word Problem/Trig, Chapter 14

15. Word Problem/Trig, Chapter 14

16. Parallel Lines on Coordinate Plane/Slope Formula, Chapter 3

17. Triangle on Coordinate Plane/Pythagorean Theorem, Chapter 8

18. Volume of Cylinder, Chapter 10

Just as we saw on the Practice PBA earlier, there are no Chapter 15 questions, but there are plenty of Chapter 12 and 14 questions -- dilations/similarity and trigonometry. Comparing this list to what has appeared on the blog so far this year, we see that a Chapter 3 question on slope appears. This year, just like last year, I'm saving slope for Chapter 11 along with more coordinate geometry. What makes that question odd is that to me, it's much easier to use

*translations*to solve the problem rather than the Slope Formula.

The question that stands out as a huge omission from the blog is Dividing Into Parts. We know that dividing a segment into parts is a major Common Core topic. Usually, the segment is given on the coordinate plane. But in this question, the segment is actually on the

*one-dimensional number line*, with the student asked to provide the coordinate of the

*point*that will divide the segment into the desired ratio. Therefore, we actually could have given this question as early as Chapter 1! Indeed, it fits very well with Lesson 1-8 on One-Dimensional Figures. And with the students already seeing such part-of-the-way questions in Chapter 1, they might be less frightened by them when they see the two-dimensional versions later on.

Now let's look at the chapter correspondence for the EOY:

1. Area of a Sector, Chapter 8

2. Octagon Reflection/Rotational Symmetry, Chapter 5

3. Dilations, Chapter 12

4. Equation of a Circle, Chapter 11

5. Reflections/Rotations in Coordinate Plane, Chapter 6

6. Height of Rectangle/Similar Triangles, Chapter 14

7. Proof of Circle Area Formula, Chapter 8

8. Equation of a Circle, Chapter 11

9. Similar Triangles, Chapter 12

10. Dilations, Chapter 12

11. Solids of Revolution, Chapter 9

12. Dividing Into Parts (2D coordinate plane), Chapter 11

13. Area of Rectangle/Population Density, Chapter 8

14. Transformations, Chapter 6

15. Volume of Box and Cylinder, Chapter 10

16. Dilations, Chapter 12

17. Translations on Coordinate Plane, Chapter 6

18. Similar Triangles Proof, Chapter 12

19. Dividing Into Parts (2D coordinate plane), Chapter 11

20. Inscribed Angle Theorem, Chapter 15

21. Basic Constructions, Chapter 3

22. Volume of Cylinder ans Sphere, Chapter 10

23. Translations and Glide Reflections, Chapter 6

24. Area of a Circle, Chapter 8

25. Triangle Constructions, Chapter 4

26. Basic Measurements, Chapter 3

27. Pythagorean Theorem/Similar Triangles, Chapter 14

28. Sines/Trig, Chapter 14

29. Word Problem/Trig, Chapter 14

30. Circle Angle Formulas, Chapter 15

31. Proof of Isosceles Triangle Theorem, Chapter 5

32. Circle Angle Formulas, Chapter 15

33. Transformations on Coordinate Plane, Chapter 6

34. Proof of Triangle-Sum Formula, Chapter 5

Let's look at some of these in more detail. The questions that appear in chapters that we have already passed are on constructions. Question 21 has the students construct first a perpendicular bisector, and then parallel lines via the Corresponding Angles Test. On the blog, I proposed a way to construct parallel lines without requiring the Corresponding Angles Test. According to the schedule, the Parallel Tests are next week, so then we can perform this construction.

Question 25 contains an interesting construction -- that of an isosceles triangle. The students must identify the conditions when the triangle is isosceles and when it is equilateral -- and the equilateral triangle construction is of course Euclid's construction of Lesson 4-4. It may be argued that the exact conditions for an isosceles triangle -- that the compass must be set to more than half the length of the given base -- follows from the Triangle Inequality, which we are far away from proving.

We are currently in the unit on symmetrical polygons and translations. So let's take a look at the questions that are relevant to the current unit:

2. Octagon Reflection/Rotational Symmetry, Chapter 5

5. Reflections/Rotations in Coordinate Plane, Chapter 6

14. Transformations on Coordinate Plane, Chapter 6

17. Translations, Chapter 6

23. Translations and Glide Reflections, Chapter 6

31. Proof of Isosceles Triangle Theorem, Chapter 5

33. Transformations on Coordinate Plane, Chapter 6

34. Proof of Triangle-Sum Theorem, Chapter 5

Aha, so we see that there's a proof of the Isosceles Triangle Theorem. We just covered this theorem here on the blog on Monday. Let's look at the PARCC proof -- the one given in the provided answer key -- so that we may compare it to the proof we gave on Monday:

Given: In Triangle

*ABC*shown,

*=*~~BA~~

~~BC~~

Prove: Angle

*A*=

*C*

*Proof:*

Statements Reasons

1.

*=*~~BA~~

2. Let

*D*be the midpoint of

3.

4.

5. Triangle

*ADB*=

*CDB*5. SSS Congruence Postulate

6. Angle

*A*=

*C*6. CPCTC

We see that this is clearly not the same as the Pappas proof that I posted on Monday. I posted that proof because I thought that it best helped students see that an isosceles triangle is symmetrical -- that is, that it can be reflected upon itself.

But look at the two problems on the PARCC PBA and EOY that do pertain to polygon symmetry -- one refers to a regular hexagon, the other to a regular octagon. So we see that the U of Chicago text spends so much time on the symmetries of isosceles triangles and kites, while these don't appear in the released PARCC questions. On the other hand, Lesson 7-6 discusses regular polygons, but there is zero mention of the rotations and reflections mapping the regular polygons to themselves.

We look at the relevant Common Core Standard again:

CCSS.MATH.CONTENT.HSG.CO.A.3

Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.

And the emphasis appears to be on

*regular polygon*, not on any of the quadrilaterals mentioned here.

If we drop the Pappas proof, then all the other known proofs involve adding some sort of auxiliary line to the isosceles triangle. We see that the PARCC proof is actually drawing in the

*median*

*D*is the midpoint of the base. Then SSS is used to prove that the two parts of the isosceles triangle are congruent triangles. For simplicity, let's call this the median-SSS proof.

But this is not the only possible proof. The U of Chicago's Isosceles Triangle Symmetry Theorem begins by drawing an angle bisector. It doesn't depend on triangle congruence, but we see that the two triangles into which the angle bisector divides the isosceles triangle are congruent by SAS. Dr. Franklin Mason also uses this proof -- which we'll call the angle bisector-SAS proof. (Indeed, as we've seen so many times, Dr. M can't follow PARCC and use SSS to prove the Isosceles Triangle Theorem because it would be circular, for he uses the ITT to prove SSS!)

And there's yet a third proof. The auxiliary line we draw can be the

*altitude*from point

*B*. Then the two resulting parts of the isosceles triangle are congruent by HL! Admittedly this is the rarest proof of them all, but that now gives us

*three*possible proofs involving auxiliary lines -- median-SSS, angle bisector-SAS, and altitude-HL.

So now we ask, which of these proofs should we teach? We might say that the PARCC encourages the median-SSS proof. Indeed, we see that the PARCC places a special emphasis on medians -- even though in theory any of the concurrency theorems can appear on the PARCC, only the concurrency of the medians -- the centroid -- actually appears on any released practice questions.

But who is the PARCC to dictate which ITT proof to use? Of course,

*some*proof of the ITT has to appear on the PARCC, and so students should be ready to give whichever proof any test expects them to give. It's most important that students simply know

*any*proof of the ITT -- again, our focus should be on which the students understand most easily. So if a standardized test like the PARCC has to ask for any proof, it should be the one that is best for the students.

I tried to find out via a Google search which of the proofs median-SSS and angle bisector-SAS is more common, or is easier for students to understand, But it's inclusive -- I see both proofs at numerous websites. On the other hand, altitude-HL doesn't really appear much at all.

Notice that just as there are three ways to prove the Isosceles Triangle Theorem using an auxiliary line, there are two ways to prove its converse. The U of Chicago text uses angle bisector-AAS, while it's also possible to use altitude-AAS -- both methods use AAS because the congruent sides that we get for free (using the reflexive property) are always opposite the given congruent angles. But we can't use the medians because this leads to SSA, and it's impossible to convert this SSA to a valid congruence such as HL or SsA.

Our Pappas-like proof of Converse Isosceles Triangle uses ASA,Meanwhile, we can't give any proof using AAS at this time because I haven't given AAS yet. AAS is problematic because the proof given in most texts is not neutral, and to make it neutral we need something like the Triangle Exterior Angle Inequality, which we choose not to use.

I was thinking back to August when I posted Legendre's spherical geometry -- a geometry in which Converse Isosceles Triangle holds, yet AAS doesn't hold. Legendre was able to prove the theorem indirectly using forward Isosceles Triangle plus a uniqueness theorem -- and this is a common strategy of proving converses. The uniqueness theorem he uses is SAS. In short, there can't be a scalene triangle with congruent base angles because there already exists an isosceles triangle with the same base angles that is congruent to the original triangle by SAS, a contradiction. I've avoided this proof in my Geometry course because I didn't want to cover indirect proofs so early, but we've already opened that can of worms this year.

Let's look at the other PARCC questions corresponding to Chapters 5 and 6. Question 34 asks students to prove the Triangle-Sum Theorem. The proof is almost the same as the one given in Lesson 5-7 of the U of Chicago text. The main difference is that the U of Chicago simply using Playfair to draw the auxiliary line

*BD*parallel to

*AC*, while the text sets up

*BD*using the Angle Measure (Protractor) Postulate and then proves that

*BD*| |

*AC*, since Playfair doesn't appear in the text until Chapter 13. Also, the U of Chicago text has separate steps for Angle Addition Postulate and the Linear Pair Theorem, but the PARCC combines these both into a single Angle Addition step.

This leaves us with four transformation questions, all on the coordinate plane. Of the transformations that appear, the reflections are both over the coordinate axes and translations are always easy, but both rotations have centers other than the origin. In particular, both of them ask students to rotate triangles

*ABC*a multiple of 90 degrees about point

*A*-- so at least we know that the image of

*A*is

*A*itself.

I'm still trying to decide how best to teach transformations on the coordinate plane. I almost want to give up -- just teach Geometry traditionally, then squeeze in a massive unit on the coordinate plane, including transformations, just before the PARCC. But actually, there are a few questions about transformations that don't require a coordinate plane -- including one on dilations, in addition to the regular polygon symmetry questions. Translations, on the other hand, are best understood by using a coordinate plane -- after all, translations consist of a magnitude and a direction (vector), and these are easy to perform on the coordinate plane.

Reflections about the four most common mirrors --

*x*-axis,

*y*-axis,

*y*=

*x*,

*y*= -

*x*-- are all easy to perform as well as prove. I've hinted before that the proof that the reflection of (

*a*,

*b*) over the line with equation

*y*=

*x*is (

*b*,

*a*) can be proved using Lambert quadrilaterals.

But what should we do about today's lesson? I ended up sticking to my original idea -- to rewrite Lesson 5-4, putting more emphasis on triangle congruence. It follows quite well from the Lesson 5-1 on Isosceles Triangles.

Question 9 is a review that asks for the median-SSS proof of the Isosceles Triangle Theorem. I point out that Question 9 can actually help with Question 8, which asks for a proof that a quadrilateral that has a pair of opposite congruent angles and a pair of congruent sides (both of which are included in the pair of congruent angles), then the quadrilateral is a kite. This proof isn't easy since blindly dividing the quadrilateral along one of the diagonals leads to SSA, which is invalid. So instead, we divide it along the

*other*diagonal into two triangles -- one of which is isosceles. The trick is to prove that the other must also be isosceles and so the quadrilateral is a kite. As it turns out, the Lambert quadrilateral that we eventually want to show is a kite (to help with our

*y*=

*x*reflection proof) requires the proof of Question 8.

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