52 Congruence of triangles; parallel lines Ch. 4
Then by the SAS principle,T(B,C,A)→≡(B′,C′,D′)T′′. In particular
|∠C′B′D′|◦=|∠CBA|◦=|∠C′B′A′|◦.Then we have different wedge-angles∠C′B′A′,∠C′B′D′, laid off on the same side of
B′C′and having the same degree-measure. This gives a contradiction.
Thus|C′,A′|=|C,A|, and as we also have
|C′,B′|=|C,B|,|∠B′C′A′|◦=|∠BCA|◦,by the SAS principle we haveT(B,C,A)→≡(B′,C′,A′)T′.
This is known as theASA (angle, side,angle)principle of congruence.
If T and T′are triangles with vertices{A,B,C},{A′,B′,C′}, respectively, for
which
|B,C|=|B′,C′|,|C,A|=|C′,A′|,|A,B|=|A′,B′|,
then T(A,B,C)→≡(A′,B′,C′)T′.
A
B
D
C
E
Figure 4.4. The SSS principle of congruence.A
B
D
E C
Proof. ChooseDon the opposite side
of BC fromA,sothat|∠CBD|◦=
|∠C′B′A′|◦and|B,D|=|B′,A′|.LetT′′
be the triangle with vertices{B,C,D}.
Then as|B,C|=|B′,C′|,bytheSAS
principle
T′′(B,D,C)→≡(B′,A′,C′)T′.Now |B,A|= |B′,A′| = |B,D| so
we have an isosceles triangle and
|∠BAD|◦ = |∠BDA|◦. Similarly
|∠CAD|◦=|∠CDA|◦.
A
B
D
C E
Note thatAandDare on different sides ofBC, so a pointEof[A,D]is onBC.
CASE 1. LetE∈[B,C].Then[A,D ⊂IR(|BAC)and[D,A∈IR(|BDC).It
follows that
|∠BAC|◦=|∠BAD|◦+|∠DAC|◦=|∠BDA|◦+|∠ADC|◦=|∠BDC|◦.