50 Congruence of triangles; parallel lines Ch. 4
We say thatTiscongruenttoT′, writtenT≡T′,ifTis congruent toT′in at
least one of the correspondences
(A,B,C)→ (A′,B′,C′),(A,B,C)→(A′,C′,B′),(A,B,C)→(B′,C′,A′),
(A,B,C)→ (B′,A′,C′),(A,B,C)→(C′,A′,B′),(A,B,C)→(C′,B′,A′).
COMMENT. Originally, behind the concept of congruence lay the idea thatTcan
be placed onT′, fitting it exactly.
AXIOM A 6 .If triangles T and T′, with vertices{A,B,C}and{A′,B′,C′}, respec-
tively, are such that
|C,A|=|C′,A′|,|A,B|=|A′,B′|,|∠BAC|◦=|∠B′A′C′|◦,
then T(A,B,C)→≡(A′,B′C′)T′.|
COMMENT. This is known as theSAS (side, angle, side)principle of congruence
for triangles.
Triangles have the following properties:-
If in a triangle[A,B,C]:-(i)|A,B|=|A,C|then|∠ABC|◦=|∠ACB|◦;(ii)|A,B|=|A,C|and D is the mid-point of B and C, then AD⊥BC;(iii)B=C, D is the mid-point of B and C, and A=D is such that AD⊥BC, then
|A,B|=|A,C|;(iv) |BACis not straight, E∈[A,B,F∈[A,C are such that|A,E|=|A,F|> 0 ,
and G=mp(E,F),thenAG=ml(|BAC).Proof.
(i) Note that for the triangleTwith vertices{A,B,C}, under the correspondence
(A,B,C)→(A,C,B),
|A,B|=|A,C|,|A,C|=|A,B|,|∠BAC|◦=|∠CAB|◦,
so by the SAS principleT(A,B,C)→≡(A,C,B)T. In particular|∠ABC|◦=|∠ACB|◦.