I wrote on this site that the last thing a student wants to see is algebra in a geometry class. But my student didn't have much trouble with the algebra for these proofs -- as long as it's simple algebra and not, say, the quadratic formula or graphing! He appears to be getting a little more comfortable with the comfortable with the idea of proof.

And so we finally reach Section 7-2 of the U of Chicago text, the Triangle Congruence Theorems. I will be able to demonstrate how SSS, SAS, ASA, and AAS follow from the definition of congruence in terms of isometries.

The text begins with SSS. We are given two triangles,

*ABC*and

*DEF*, with

*,*~~AB~~

*, and*~~BC~~

*congruent to*~~AC~~

*,*~~DE~~

*, and*~~EF~~

*, respectively. The text begins by noting that since*~~DF~~

*and*~~AB~~

*are congruent, there must exist some isometry mapping*~~DE~~

*to*~~AB~~

*-- after all, that is exactly what it*~~DE~~

*means*to be congruent under the Common Core definition. So now we ask, what is the whole triangle

*ABC*mapped to under this same isometry?

If we're lucky,

*C*will already be mapped to

*F*. Then this same isometry will end up mapping the entire triangle

*ABC*to

*DEF*-- which would already make the two triangles congruent. So let's say that

*ABC*is not mapped to

*DEF*. We already mapped

*to*~~AB~~

*-- that is, we already mapped*~~DE~~

*A*to

*D*and

*B*to

*E*. So

*C*is not mapped to

*F*, but to some new point,

*C'*.

The text now looks at the quadrilateral

*DFEC'*and notes that it satisfies the definition of one of our special quadrilaterals -- the

*kite*. By the Kite Symmetry Theorem from Chapter 5, the kite contains a symmetry diagonal,

*. This means that the reflection over the line*~~DE~~

*DE*must map half of the kite, triangle

*DEC'*, to the other half,

*DEF*. Since

*ABC*is congruent to

*DEC'*(as the original isometry maps the former to the latter) and

*DEC'*is congruent to

*DEF*(as the reflection over line

*DE*maps the former to the latter), by the Transitive Property,

*ABC*is congruent to

*DEF*. QED

The text moves on to SAS and ASA. These proofs are similar. In each case, we begin with an isometry mapping one of the congruent sides to the other -- without loss of generality, we call these congruent sides

*and*~~AB~~

*. In the SAS case, if*~~DE~~

*C'*is not already

*F*, then

*C'DF*is an isosceles triangle, and so the Isosceles Triangle Symmetry Theorem in this proof guarantees that the reflection image of

*C'*over line

*DE*-- the line containing the bisector of angle

*C'DF*-- is

*F*(just as the Kite Symmetry Theorem did in the SSS proof). For ASA, line

*DE*now bisects two angles,

*C'DF*and

*C'EF*. The text then uses the Side-Switching Theorem to show that the reflection images of rays

*DC'*and

*EC'*must be

*DF*and

*EF*respectively, forcing

*C'*to reflect to

*F*once again. Both proofs end the same way as the SSS proof --

*DEC'*reflects to

*DEF*, so they must be congruent to each other and thus to

*ABC*.

In all three proofs, the trick is to map

*ABC*to a triangle that is, if not

*DEF*itself, then a triangle that reflects to

*DEF*. In no case is the isometry mapping

*ABC*to

*DEC'*made explicit -- only that some isometry must exist. Only the isometry from

*DEC'*to

*DEF*is ever made explicit -- this isometry is always a reflection.

I have much more to say about the proofs of SSS, SAS, and ASA -- but once again, I don't have much time to write it now.

Meanwhile, I'm still having problems figuring things out on the new computer. I have my printer scanner from the old computer. I can send documents to the printer to print, but I just can't get the computer to receive documents from the scanner. This is such an important lesson that I simply must have these images posted to this blog. Hopefully at some point I can finally have time to do things, such as figure out how to get documents to scan properly.

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