# Congruence Criteria for Triangles—AAS and HL

### New York State Common Core Math Geometry, Module 1, Lesson 25

Worksheets for Geometry, Module 1, Lesson 25

Student Outcomes

• Students learn why any two triangles that satisfy the SAA or HL congruence criteria must be congruent.
• Students learn why any two triangles that meet the AAA or SSA criteria are not necessarily congruent.

Congruence Criteria for Triangles—AAS and HL

Classwork

Opening Exercise

Write a proof for the following question. Once done, compare your proof with a neighbor’s. Given: 𝐷𝐸 = 𝐷𝐺, 𝐸𝐹 = 𝐺𝐹 Prove: 𝐷𝐹 is the angle bisector of ∠𝐸𝐷𝐺

Exploratory Challenge

Today we are going to examine three possible triangle congruence criteria, Angle-Angle-Side (AAS), Side-Side-Angle (SSA), and Angle-Angle-Angle (AAA). Ultimately, only one of the three possible criteria ensures congruence.

ANGLE-ANGLE-SIDE TRIANGLE CONGRUENCE CRITERIA (AAS): Given two triangles △ 𝐴𝐵𝐶 and △ 𝐴′𝐵′𝐶′. If 𝐴𝐵 = 𝐴′𝐵′ (Side), 𝑚∠𝐵 = 𝑚∠𝐵′ (Angle), and 𝑚∠𝐶 = 𝑚∠𝐶′ (Angle), then the triangles are congruent.

PROOF:
Consider a pair of triangles that meet the AAS criteria. 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 draw about the third angle of each triangle?

Since the first two angles are equal in measure, the third angles must also be equal in measure.

Given this conclusion, which formerly learned triangle congruence criteria can we use to determine if the pair of triangles are congruent?

Therefore, the AAS criterion is actually an extension of the ____ triangle congruence criterion. Note that when using the Angle-Angle-Side triangle congruence criteria as a reason in a proof, you need only state the congruence and AAS.

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.

PROOF:
As with some of our other proofs, we do not start at the very beginning, but imagine that a congruence exists so that triangles have been brought together such that 𝐴 = 𝐴 ′ and 𝐶 = 𝐶′; the hypotenuse acts as a common side to the transformed triangles.

Similar to the proof for SSS, we add a construction and draw 𝐵𝐵′. △ 𝐴𝐵𝐵′ is isosceles by definition, and we can conclude that base angles 𝑚∠𝐴𝐵𝐵 ′ = 𝑚∠𝐴𝐵′𝐵. Since ∠𝐶𝐵𝐵 ′ and ∠𝐶𝐵 ′𝐵 are both the complements of equal angle measures (∠𝐴𝐵𝐵 ′ and ∠𝐴𝐵′𝐵), they too are equal in measure. Furthermore, since 𝑚∠𝐶𝐵𝐵 ′ = 𝑚∠𝐶𝐵 ′𝐵, the sides of △ 𝐶𝐵𝐵 ′opposite them are equal in measure: 𝐵𝐶 = 𝐵′𝐶′.

Then, by SSS, we can conclude △ 𝐴𝐵𝐶 ≅△ 𝐴′𝐵′𝐶′. Note that when using the Hypotenuse-Leg triangle congruence criteria as a reason in a proof, you need only to state the congruence and HL.

Criteria that do not determine two triangles as congruent: SSA and AAA

SIDE-SIDE-ANGLE (SSA): Observe the diagrams below. Each triangle has a set of adjacent sides of measures 11 and 9, as well as the non-included angle of 23°. Yet, the triangles are not congruent.

Examine the composite made of both triangles. The sides of length 9 each have been dashed to show their possible locations. The triangles that satisfy the conditions of SSA cannot guarantee congruence criteria. In other words, two triangles under SSA criteria may or may not be congruent; therefore, we cannot categorize SSA as congruence criterion

ANGLE-ANGLE-ANGLE (AAA): A correspondence exists between △ 𝐴𝐵𝐶 and △ 𝐷𝐸𝐹. Trace △ 𝐴𝐵𝐶 onto patty paper, and line up corresponding vertices.

Based on your observations, why isn’t AAA categorizes as congruence criteria? Is there any situation in which AAA does guarantee congruence?

Even though the angle measures may be the same, the sides can be proportionally larger; you can have similar triangles in addition to a congruent triangle.

List all the triangle congruence criteria here: ___________

List the criteria that do not determine congruence here: ________

Examples

1. Given: 𝐵𝐶 ⊥ 𝐶𝐷, 𝐴𝐵 ⊥ 𝐴𝐷 , 𝑚∠1 = 𝑚∠2
Prove: △ 𝐵𝐶𝐷 ≅△ 𝐵𝐴𝐷

2. Given: 𝐴𝐷 ⊥ 𝐵𝐷, 𝐵𝐷 ⊥ 𝐵𝐶, 𝐴𝐵 = 𝐶𝐷
Prove: △ 𝐴𝐵𝐷 ≅△ 𝐶𝐷𝐵

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