Sec. 4.1 Principles of congruence 51
(ii) Note that if T 1 ,T 2 are
the triangles with vertices
{A,B,D},{A,C,D}, respec-
tively, then
|A,B|=|A,C|,|B,D|=|C,D|,|∠ABD|◦=|∠ACD|◦,by(i),so by the SAS principle,
T (^1) (A,B,D)→≡(A,C,D) T 2 .Inpar-
ticular |∠ADB|◦ = |∠ADC|◦.
AsD∈[B,C], the sum of the
degree-measures of these angles
is 180 and so they must be
right-angles.
A
BDC
Figure 4.2. An isosceles triangle.(iii) AsAD⊥BCwe know thatA∈BC.IfT 1 ,T 2 are the triangles with vertices
{A,B,D},{A,C,D}, respectively, then
|B,D|=|C,D|,|A,D|=|A,D|,|∠BDA|◦=|∠CDA|◦,so by the SAS principle,T (^1) (A,B,D)→≡(A,C,D)T 2. In particular|A,B|=|A,C|.
(iv) As in (ii), the triangles[A,E,G],[A,F,G]are congruent, and so
|∠EAG|◦=|∠FAG|◦.
Definition. A triangle is said to beisoscelesif at least two of its sides have equal
lengths.
If T,T′are triangles with vertices{A,B,C},{A′,B′,C′}, respectively, for which
|B,C|=|B′,C′|,|∠CBA|◦=|∠C′B′A′|◦,|∠BCA|◦=|∠B′C′A′|◦,
then T(A,B,C)→≡(A′,B′,C′)T′.
Proof. Suppose that|C′,A′|=|C,A|. Choose the pointD′on the half-line[C′,A′
such that|C′,D′|=|C,A|.ThenifT′′is the triangle with vertices{B′,C′,D′}, under
the correspondence(B,C,A)→(B′,C′,D′)we have
|B,C|=|B′,C′|,|C,A|=|C′,D′|,|∠BCA|◦=|∠B′C′D′|◦.
A
B
C
A′
B′
C′
D′
Figure 4.3.