With all the stuff in the high school geometry about proving congruence by rigid motions we get this sample geometry question from PARCC

(get the rest from numberwarrior here)

Numberwarrior’s concerns are about the language and the formal properties of congruence and I agree with him on this.

My concerns are about the stated claims of the CCSS to specify the “What do the need to know/understand/be able to do”, and the PARCC test which says “This is HOW you do a proof”.

In this particular example there are other ways of proving the assertion, not least those using the definition of congruence by rigid motions.

Let us do it this way:

1: vertical angles are equal, as there is a rotation of line AD to GC through the angle CBD, and then AD is on top of GC, so angle ~~ABD~~ ABF is also the angle of rotation, and is therefore congruent to angle CBD

2:There is a translation of line HE to line AD, as they are parallel. So the translation of H to H’ puts H’ on the line AD, and so angle H’BF is congruent to angle ABF.

3: But angles are preserved by rigid motions, so angle H’BF is congruent to HFG, and therefore angle ABF and HFG are congruent.

So, if I chose to teach about proof using this approach (my “HOW”) the students won’t even understand the question. Test items MUST be “Method Free”.

Also, the so called Reflexive, Symmetric and Transitive properties of congruence are no different from a=a, if a=b then b=a, and if a=b and b=c then a=c for numbers, and in both situations these are so STUNNINGLY obvious that it is cluttering up the minds of the learners to burden them with this sort of stuff. It is clear to me that this is a contribution to the CCSS from the sole pure mathematician on the committee.

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Let’s hope PARCC is listening?

Step 1: “… so angle ABD is also the angle of rotation, and is therefore congruent to angle CBD.” Did you mean angle ABF?

Step 2: Do we know, at this point, that lines HE and AD are parallel?

“… so STUNNINGLY obvious that it is cluttering up the minds of the learners.” Yup.

Thanks, Howard.

To Fawn:

Yes, I did mean ABF, not ABD. It was late! I am going to fix it now.

About step 2, Pythagoras says that if two lines cross a third line and they make different angles with that line then the two lines intersect. In the (translated) account I read recently he doesn’t use the word parallel at that point. So the two lines in the problem are parallel. What needs to be specified is the definition of “parallel”, and there are several!

There is a much neater way of looking at this, which a beginner is very unlikely to spot, and that is to see the picture as two potentially congruent figures. The first is the two lines AD and CG, the second is the two lines HE and CG (yes, CG is in BOTH figures). Then a translation of the second to the first carrying F to B does it in one step.

If I can find a mailing address for PARCC I will send them this.

And many thanks for your comment.