Sign In Sign-Up

Welcome!

Close

Would you like to make this site your homepage? It's fast and easy...

Yes, Please make this my home page!

No Thanks

  Don't show this to me again.

Close

Jeremy Gibbs

Modern Geometry

10-13-97

Proof of Proposition 3.22

Proposition 3.22 (SSS Criterion for Congruence). Given triangle ABC and triangle DEF. If AB = DE, BC = EF, and AC = DF, then triangle ABC = triangle DEF.

Outline for proof

I. Given the two triangles, ABC and DEF and that AB = DE, BC = EF, and AC = DF.

II. By the corollary to SAS, given the triangle ABC and segment DE = AB, there is a unique point F on a given side of line DE such that triangle ABC = triangle DEF.

III. Apply the corollary again to deduce that given the triangle ABC and segment BC = EF, then there is an unique point D on a given side of line EF such that triangle DEF = triangle ABC.

IV. Apply the corollary once again to deduce that given the triangle ABC and segment CA = FD, then there is an unique point E’ on a given side of line FD, such that triangle DE’F = triangle ABC.

V. By the corollary to CA 6, we can show that triangle ACE’ = triangle DEF. Since A = D, C = F, then we can show that E’ = E since triangle ACE’ is just a copy of triangle DEF, only placed so that it is connected to triangle ABC.

VI. We are left with the fact that A = D, C = F, and the points B and E’ are on opposite sides of line AC.

VII. There are three cases to consider: A*G*C; A=G C = G; or G*A*C, A*C*G.

VIII. If A*G*C, then we can see by Proposition 3.10 that since AB = AE, then the angles <ABE = <AEB. Also, by the same proposition, angles <CBE = <CEB. By Proposition 3.19, which allows for angle addition, we see that <ABC = <AEC.

IX. Then, by CA 6 (SAS), we show that since AB = AE, BC = CE, and <CBE = <CEB, then triangle ABC = triangle AEC (DEF).

X. Second case: G is equal to either A or C. Then, by Proposition 3.10 we see that <CBA = <CEA since BC = EC. Thus, by CA 6 triangle ABC = triangle AEC (DEF).

XI. Third case: G*A*C or A*C*G. We first use Proposition 3.10 to see that <AEG = < ABG. Next, we use Proposition 3.10 to show that <CEG = <CBG. Using Proposition 3.20 which allows us to subtract angles, we show that <CBG - <ABG = <CEG - <AEG or <CBA = <CEA. Using CA 6, we see that triangle ABC = triangle AEC (DEF).

Given two triangles, ABC and DEF and that AB = DE, BC = EF, and AC = DF. Using the corollary to SAS, given the triangle ABC and segment DE = AB, there is a unique point F on a given side of line DE such that triangle ABC = triangle DEF. Applying the corollary again, we see that using the given of triangle ABC and segment BC = EF, then there is a unique point D on a given side of line EF such that triangle DEF = triangle ABC. Applying the corollary once again, we see that using the given of triangle ABC and CA = FD, then there is a unique point E’ on a given side of line FD, such that triangle DE’F = triangle ABC. Thus, we are left with the fact that A = D, C = F, and the points B and E’ are on opposite sides of line AC. Since that is true, we can show that triangle ACE’ = triangle DEF since A = D, C = F, and E = E’ since by the Corollary to CA 6, it is just a copy of the triangle DEF. From henceforth, we will just refer to triangle ACE’, or DEF, as ACE. Now, we have three cases to consider. We will call G the point where segment BE meets segment AC. The cases are: A*G*C, A=G or C=G, and G*A*C or A*C*G. If A*G*C, then we can see by Proposition 3.10 that since AB = AE, then the angles <ABE = <AEB. Also, by the same propositoin, angles <CBE = <CEB. By Proposition 3.19, which allows for angle addition, we see that <ABC = <AEC. Then, by CA 6 (SAS), we see that since AB = AE, BC = CE, and <CBE = <CEB, then triangle ABC = triangle AEC, which we will call DEF since A = D and C = F. Thus, case one is satisfied. The second case involves the point G being equal to either A or C. Then, by Proposition 3.10, we see that <CBA = <CEA since BC = EC. Thus, we can show that since BC = CE, AB = AE, and angle <CBA = <CEA, then by CA 6, triangle ABC = triangle AEC (DEF). Thus, the second case is satisfied. The third case to consider is when G is not in between A and C. So, either, G*A*C or A*C*G. We first will use Proposition 3.10 to see that <AEG = <ABG. Next, we use Proposition 3.10 again to show that <CEG = <CBG. Using Proposition 3.20 which allows us to subtract angles, we show that <CBG - <ABG = <CEG - <AEG or <CBA = <CEA. Using CA 6, we see that since AB = AE, BC = EC, and <CBA = <CEA, then triangle ABC = triangle AEC or DEF. Thus, since all three cases have been considered and proven correctly, the proof is complete showing that if two triangles have three sides that match up to be congruent, the triangles themselves will also be congruent.