Proposition 4: SAS Triangle Congruence

Transcription

1 Proposition 4: SAS Triangle Congruence The method of proof used in this proposition is sometimes called "superposition." It apparently is not a method that Euclid prefers since he so rarely uses it, only here in I.4 and in I.8 and III.24, but not in many other propositions in which he could have used it. It is not entirely clear what is meant by the statement "triangle ABC is applied to triangle DEF" means. It has been variously interpreted as actually moving one triangle to cover the other or as simply associating parts of one triangle with parts of the other. For the two triangles illustrated in the figure, you can actually slide one over the other in a continuous motion within the plane. Note, however, that if one triangle is the mirror image of the other, then any continuous motion would require moving one triangle outside of the plane. But the triangles don't have to be in the same plane to begin with, and they often are not in the same plane when this proposition is invoked in the books on solid geometry. Whatever the intended meaning of superposition (i.e., applying one triangle to another) may be, there are no postulates to allow any conclusions based on superposition. One possibility is to add postulates based on a group of transformations of space, or if restricted to plane geometry, on a group of transformations of the plane. Charles Dodgson (a.k.a. Lewis Carroll) would have said that using group theory is not appropriate to an elementary exposition of Euclidean geometry. The mathematics historian, Sir Thomas Heath, has described a more elementary conservative basis in his commentary on this proposition. Yet another alternative is to simply take this proposition as a postulate, or part of it as a postulate. For instance, David Hilbert in his Foundations of Geometry takes as given that under the hypotheses of this proposition that the remaining angles equal the remaining angles. Then, Hilbert proves that the base equals the base.

2 On The Congruence Of Triangles by Jim Loy In Plane Geometry, two objects are congruent if all of their corresponding parts are congruent. In the first diagram, the two triangles have two sides which are congruent, and the angle between these sides is also congruent. Euclid proved that they are congruent triangles (Theorem I.4, called "Side Angle Side" of SAS). But, he was not happy with the proof, as he avoided similar proofs in other situations. The way he proved it, is to move one triangle until it is superimposed on the other triangle. Such a trick (superposition: placing one triangle on top of another, to see if they are congruent), is not considered legal, now. It involves some complicated assumptions. So, now this (SAS) is given as an assumption (postulate). It is the Side Angle Side Postulate. Euclid proved his Side Side Side (SSS) Theorem (I.8) and his Angle Side Angle (ASA, diagram at the right) Theorem (I.26) in a similar way. In SSS, if a triangle has all three sides congruent to the corresponding sides of a second triangle, then they are congruent. In ASA, if a triangle has two angles and the side in between the angles congruent to the corresponding parts of a second triangle, then they are congruent. These too, are now assumed as postulates. It is easy to construct a triangle given the lengths of the three sides (diagram on the left), or two sides and the included angle, or two angles and the included side. Given the ease of these constructions (especially SSS), it seems strange that the corresponding congruence "theorems" cannot be proven, except by slightly shady means (superposition). All three of these congruence postulates are equivalent. You can assume any one of them, and prove the other two from there. Euclid also proved an Angle Angle Side (AAS) as a corollary (a minor theorem derived directly from the parent theorem) to Theorem I.26. The situation SSA (or ASS) is not a theorem or postulate, as it is often ambiguous, there are often two different non congruent triangles with two congruent sides and a congruent angle at the end of one of the sides.

3 What follows is a proof that all three congruence criteria are equivalent in four parts: 1. Assume SAS and prove ASA. ;A =;EDF, ;B =;E, and AB = DE. 1. If BC = EF, then the two triangles are congruent, by SAS. So, assume that BC EF. 2. There is a point G, on EF (perhaps extended), on the same side of the line DE, so that EG = BC. 3. Δ ABC Δ DEG [by SAS]. 4. ;EDG = ;A [definition of congruent triangles]. 5. ;EDF = ;EDG [both =;A ]. 6. So DF and DG are the same line. 7. DF and EF intersect at both F and G. So, F & G are the same point [two distinct lines intersect in only one point]. 8. So EF = EG = BC, which contradicts our assumption that BC EF. 9. So BC = EF, and the triangles are congruent by SAS, as mentioned in step #1. 2. Assume ASA and prove SAS. ;A =;D, AB = DE, and AC = DF. 1. If ;B = ;DEF, then the two triangles are congruent, by ASA. So, assume that ;B ;DEF. 2. Draw a line EG, so that ;DEG =;B (with G on the same side of DE as F). 3. EG intersects line DF (perhaps extended) at some point H. 4. Δ ABC ΔDEH [ASA]. 5. AC = DH [definition of congruent triangles]. 6. DH = DF [both = AC ]. 7. H and F are the same point. And EH and EF are the same line. And ;DEG and ;DEF are the same angle. 8. That contradicts our assumption that ;B ;DEF. 9. So ;B = ;DEF, and the two triangles are congruent by ASA, as mentioned in step #1.

4 3. Assume SSS and prove SAS. ;A=;EDF, AB = DE, and AC = DF. 1. If BC = EF, then the two triangles are congruent, by SSS. So, assume that BC EF. 2. Find the point G (on the same side of line DE as F is) so that DF = DG and BC = EG. We do this by drawing arcs of those radii and finding where they intersect, if they do intersect. 3. Δ ABC Δ DEG [SSS]. 4. ; A = ;EDG [definition of congruence of triangles]. 5. ;EDG = ;EDF [both=;a]. 6. So, they are the same angle. And F and G are the same point. And EF = BC. 7. This contradicts our assumption that BC EF. 8. So BC = EF, and the triangles are congruent by SSS, as mentioned in step #1. 4. Assume SAS and prove SSS. AB = DE, AC = DF, and BC = EF. 1. If ;A = ;EDF, then the two triangles are congruent, by SAS. So, assume that ;A ;EDF. 2. Draw a line DG, so that ;BAC = ;EDG. 3. Find the point H on DG (on the same side of DE that F is) so that DH = AC. 4. F & H are not the same point [DF and DG can only intersect at D]. 5. Draw EH and FH. 6. Δ ABC ΔDEH [SAS]. 7. EH = BC [definition of congruent triangles]. 8. EH = EF [both = BC ]. 9. DH = DF [both = AC ]. 10. Both ΔDFH and ΔEFH are isosceles [definition]. 11. The perpendicular bisector of FH must pass through both D and E [a Euclid theorem, which is provable using the SAS congruency criterion]. 12. So DE is the perpendicular bisector of FH [definition]. 13. But that is impossible, since H and F are on the same side of line DE [step #3]. 14. This contradicts our assumption that ; A ;EDF. 15. So ;A =;EDF, and the triangles are congruent by SAS, as mentioned in step #1.

5 ASA ASA SAS SSS SAS SSS Note: Above I showed that ASA implies SAS and vice versa. I also showed that SAS implies SSS and vice versa. So, you can assume any of the three, and then go by a chain of above proofs to prove any of the other two. See the left half of this diagram. This kind of chain can be done in three proofs, rather than four. See the right half of the diagram, to see how that could be done. I had difficulty coming up with any proof involving ASA and SSS. So, I did it with four proofs. Notice that all four of the above proofs are fairly similar. That is further evidence that these three theorems/postulates are equivalent. And none of them is much more obvious than any of the others.