Download GEOMETRY MODULE 1 LESSON 25

GEOMETRY
MODULE 1 LESSON 25
CONGRUENECE CRITERIA FOR TRIANGLES –AAS and HL
OPENING EXERCISE
Write a proof for the following question. You may work in groups.
Given: 𝐷𝐸 = 𝐷𝐺, 𝐸𝐹 = 𝐺𝐹
Prove: ̅̅̅̅
𝐷𝐹 is the angle bisector of ∠𝐸𝐷𝐺.
STEP
JUSTIFICATION
1
𝐷𝐸 = 𝐷𝐺
Given
2
𝐸𝐹 = 𝐺𝐹
Given
3
𝐷𝐹 = 𝐷𝐹
Reflexive Property
4
∆𝐷𝐸𝐹 ≅ ∆𝐷𝐺𝐹
SSS
5
∠𝐸𝐷𝐹 ≅ ∠𝐺𝐷𝐹
Property of Congruent Triangles
6
̅̅̅̅ is the angle bisector of ∠𝐸𝐷𝐺
𝐷𝐹
Definition of Angle Bisector
DISCUSSION
Today, we consider three possible triangle congruence criteria: AAS, SSA, AAA.

Angle- Angle-Side Triangle Congruence Criteria (AAS): Given two triangles
∆𝐴𝐵𝐶 and ∆𝐴′𝐵′𝐶′. If 𝑚∠𝐵 = 𝑚∠𝐵′ (Angle), 𝑚∠𝐶 = 𝑚∠𝐶′ (Angle), and 𝐴𝐵 = 𝐴’𝐵’
(Side), then the triangles are congruent.
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If you knew that two angles of one triangle corresponded to and were equal in measure to two
angles of the other triangle, what conclusions can you make about the third angle of each
triangle?
The AAS criterion is actually an extension of the ASA triangle congruence criterion.
Do SSA and AAA guarantee triangle congruence?
SSA?
Observe the triangles below. Each triangle has a set of adjacent sides and a non-included angle
of 23°. Yet, the triangles are not congruent.
AAA?
In Problem Set Lesson 20, a similar question was posed. The triangles below have
corresponding angles of 60°. Yet, the triangles are not congruent.
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
Hypotenuse-Leg Triangle Congruence Criteria (HL): Given two right
triangles ∆𝐴𝐵𝐶 and ∆𝐴′𝐵′𝐶′ with right angles 𝐵 and 𝐵′. If 𝐴𝐵 = 𝐴′𝐵′
(Leg) and 𝐴𝐶 = 𝐴′𝐶′ (Hypotenuse), then the triangles are congruent.
Imagine that a congruence exists so that triangles have been brought
together such that 𝐴 = 𝐴′ (common vertex) and 𝐶 = 𝐶′ (common vertex). The hypotenuse AC
̅̅̅̅̅.
acts as a common side. We will add a construction and draw 𝐵𝐵′
STEP
1
𝐴𝐵 = 𝐴′𝐵′
JUSTIFICATION
Given
∠𝐵 = ∠𝐵 ′ = 90°
2
𝐴𝐶 = 𝐴𝐶
Reflexive Property
3
∆𝐴𝐵𝐵′ is isosceles
Definition of Isosceles
4
𝑚∠𝐴𝐵𝐵 ′ = 𝑚∠𝐴𝐵′𝐵
The measures of base angles of isosceles triangles
are equal.
5
∠𝐶𝐵𝐵 ′ + ∠𝐴𝐵𝐵 ′ = 90°
Angle Addition
∠𝐶𝐵′𝐵 + ∠𝐴𝐵′𝐵 = 90°
6
∠𝐶𝐵𝐵 ′ = 90° − ∠𝐴𝐵𝐵 ′
Subtraction
∠𝐶𝐵′𝐵 = 90° − ∠𝐴𝐵′𝐵
7
𝑚∠𝐶𝐵𝐵 ′ = 𝑚∠𝐶𝐵′𝐵
Substitution from Line 4/Transitive
8
∆𝐵𝐶𝐵′ is isosceles
Definition of Isosceles (Line 7 Base Angles)
9
𝐵𝐶 = 𝐵′𝐶′
Property of Isosceles
10
∆𝐴𝐵𝐶 ≅ ∆𝐴′𝐵′𝐶′
SSS (Line 1, 2, 9)
We have all the triangle congruencies now: SSS, SAS, ASA, AAS, HL.
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PRACTICE
̅̅̅̅ ⊥ 𝐶𝐷
̅̅̅̅, 𝐴𝐵
̅̅̅̅ ⊥ 𝐴𝐷
̅̅̅̅, 𝑚∠1 = 𝑚∠2
Given: 𝐵𝐶
Prove: ∆𝐵𝐶𝐷 ≅ ∆𝐵𝐴𝐷
STEP
̅̅̅̅ ⊥ 𝐶𝐷
̅̅̅̅ ,
𝐵𝐶
1
̅̅̅̅ ⊥ 𝐴𝐷
̅̅̅̅ ,
𝐴𝐵
JUSTIFICATION
𝑚∠1 = 𝑚∠2
Given
2
𝐵𝐷 = 𝐵𝐷
Reflexive Property
3
𝑚∠1 + 𝑚∠𝐶𝐷𝐵 = 180°
Liner pairs are supplementary.
4
𝑚∠2 + 𝑚∠𝐴𝐷𝐵 = 180°
Liner pairs are supplementary.
5
𝑚∠𝐶𝐷𝐵 = 𝑚∠𝐴𝐷𝐵
If two angles are equal in measure, then their
supplements are equal in measure.
6
𝑚∠𝐵𝐶𝐷 = 𝑚∠𝐵𝐴𝐷 = 90°
Definition of ⊥ line segments
7
∆𝐵𝐶𝐷 ≅ ∆𝐵𝐴𝐷
AAS
̅̅̅̅ , 𝐴𝐵 = 𝐶𝐷
Given: ̅̅̅̅
𝐴𝐷 ⊥ ̅̅̅̅
𝐵𝐷, ̅̅̅̅
𝐵𝐷 ⊥ 𝐵𝐶
Prove: ∆𝐴𝐵𝐷 ≅ ∆𝐶𝐷𝐵
STEP
1
̅̅̅̅
𝐴𝐷 ⊥ ̅̅̅̅
𝐵𝐷,
̅̅̅̅ ,
̅̅̅̅
𝐵𝐷 ⊥ 𝐵𝐶
JUSTIFICATION
𝐴𝐵 = 𝐶𝐷
Given
2
∆𝐴𝐵𝐷 is a right triangle.
Definition of ⊥ line segments
3
∆𝐶𝐷𝐵 is a right triangle.
Definition of ⊥ line segments
4
𝐵𝐷 = 𝐷𝐵
Reflexive Property
5
∆𝐴𝐵𝐷 ≅ ∆𝐶𝐷𝐵
HL
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