Side-Angle-Side Congruence and Isosceles Triangles in Geometry, Study notes of Geometry

The concept of congruence in geometry, specifically in the context of side-angle-side (sas) and its application to isosceles triangles. The definition of congruence, provides examples, and demonstrates the side-angle-side axiom, which states that if two triangles have corresponding sides congruent and corresponding angles congruent, then the triangles are congruent. The document also defines isosceles triangles and states that in a neutral geometry, the base angles of an isosceles triangle are congruent.

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Lecture 18: Side-Angle-Side
18.1 Congruence
Notation: Given 4ABC, we will write Afor BAC,Bfor ABC , and Cfor BCA
if the meaning is clear from the context.
Definition In a protractor geometry, we write 4ABC ' 4DE F if
AB 'DE, BC 'E F , CA 'FD ,
and
A'D, B'E, C'F.
Definition In a protractor geometry, if 4ABC ' 4DEF ,4AC B ' 4DE F ,4BAC '
4DE F ,4BCA ' 4D EF ,4CAB ' 4D EF , or 4CBA ' 4D EF , we say 4ABC and
4DEF are congruent
Example In the Taxicab Plane, let A= (0,0), B= (1,1), C= (1,1), D= (5,0),
E= (5,2), and F= (7,0). Then
AB = 2 = DE,
AC = 2 = DF,
m(A) = 90 = m(D),
m(B) = 45 = m(E),
and
m(C) = 45 = m(F),
and so AB 'DE,AC 'DF ,A'D,B'E, and C'F. However,
BC = 2 6= 4 = EF.
Thus BC and EF are not congruent, and so 4ABC and 4DEF are not congruent.
18.2 Side-angle-side
Definition A protractor geometry satisfies the Side-Angle-Side Axiom (SAS) if, given
4ABC and 4DE F ,AB 'DE,B'E, and B C 'EF imply 4ABC ' 4DE F .
Definition We call a protractor geometry satisfying the Side-Angle-Side Axiom a neutral
geometry, also called an absolute geometry.
18-1
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Lecture 18: Side-Angle-Side

18.1 Congruence

Notation: Given 4 ABC, we will write ∠A for ∠BAC, ∠B for ∠ABC, and ∠C for ∠BCA if the meaning is clear from the context.

Definition In a protractor geometry, we write 4 ABC ' 4DEF if

AB ' DE, BC ' EF , CA ' F D,

and ∠A ' ∠D, ∠B ' ∠E, ∠C ' ∠F.

Definition In a protractor geometry, if 4 ABC ' 4DEF , 4 ACB ' 4DEF , 4 BAC ' 4 DEF , 4 BCA ' 4DEF , 4 CAB ' 4DEF , or 4 CBA ' 4DEF , we say 4 ABC and 4 DEF are congruent

Example In the Taxicab Plane, let A = (0, 0), B = (− 1 , 1), C = (1, 1), D = (5, 0), E = (5, 2), and F = (7, 0). Then AB = 2 = DE,

AC = 2 = DF,

m(∠A) = 90 = m(∠D),

m(∠B) = 45 = m(∠E),

and m(∠C) = 45 = m(∠F ),

and so AB ' DE, AC ' DF , ∠A ' ∠D, ∠B ' ∠E, and ∠C ' ∠F. However,

BC = 2 6 = 4 = EF.

Thus BC and EF are not congruent, and so 4 ABC and 4 DEF are not congruent.

18.2 Side-angle-side

Definition A protractor geometry satisfies the Side-Angle-Side Axiom (SAS) if, given 4 ABC and 4 DEF , AB ' DE, ∠B ' ∠E, and BC ' EF imply 4 ABC ' 4DEF.

Definition We call a protractor geometry satisfying the Side-Angle-Side Axiom a neutral geometry, also called an absolute geometry.

Lecture 18: Side-Angle-Side 18-

Example We will show that the Euclidean Plane is a neutral geometry. First recall the Law of Cosines: Given any triangle 4 ABC in the Euclidean Plane,

AC^2 = AB^2 + BC^2 − 2(AB)(BC) cos(mE (∠B)).

Note that, in particular, the measure of any angle of a triangle in the Euclidean Plane is determined by the lengths of the sides of the triangle. Hence given 4 ABC and 4 DEF with AB ' DE, ∠B ' E, and BC ' EF , we need show only that AC ' DF to conclude that 4 ABC ' 4DEF. Now

AC^2 = AB^2 + BC^2 − 2(AB)(BC) cos(mE (∠B)) = DE^2 + EF 2 − 2(DE)(EF ) cos(mE (∠E)) = DF 2 ,

so AC = DF. Thus AC ' DF.

Example The Poincar´e Plane is a neutral geometry. We will omit the proof, which is more easily done with the help of an axiom about isometries of the plane which is equivalent to the Side-Angle-Side Axiom.

18.3 Isosceles triangles

Definition In a protractor geometry, we say a triangle with two congruent sides is isosceles. We say a triangle which is not isosceles is scalene. It 4 ABC is isosceles with AB ' BC, then we call ∠A and ∠B the base angles of 4 ABC. We say a triangle with all three sides congruent is equilateral.

Pons Asinorum In a neutral geometry, the base angles of an isosceles triangle are congruent.

Proof Consider a triangle 4 ABC with AB ' BC. Then AB ' CB, ∠ABC ' ∠CBA, and BC ' BA. Hence, by Side-Angle-Side, 4 ABC ' 4CBA. In particular, ∠BAC ' ∠BCA.