Congruent Triangles - Tenth Grade Geometry
Introduction to Triangle Congruence
Congruent Triangles: Two triangles that have exactly the same size and shape
Symbol: $\triangle ABC \cong \triangle DEF$
Key Property: All corresponding sides AND all corresponding angles are equal
Five Methods: SSS, SAS, ASA, AAS, and HL (for right triangles)
Important: You don't need to prove all 6 parts equal—certain combinations are sufficient
Symbol: $\triangle ABC \cong \triangle DEF$
Key Property: All corresponding sides AND all corresponding angles are equal
Five Methods: SSS, SAS, ASA, AAS, and HL (for right triangles)
Important: You don't need to prove all 6 parts equal—certain combinations are sufficient
1. Prove Triangles are Congruent Using Rigid Motions
Rigid Motion (Isometry): A transformation that preserves distance and angle measure
Three Types: Translation (slide), Reflection (flip), Rotation (turn)
Key Concept: Two triangles are congruent if and only if one can be mapped onto the other using rigid motions
Why It Works: Rigid motions preserve all measurements (sides and angles)
Three Types: Translation (slide), Reflection (flip), Rotation (turn)
Key Concept: Two triangles are congruent if and only if one can be mapped onto the other using rigid motions
Why It Works: Rigid motions preserve all measurements (sides and angles)
Rigid Motions and Congruence:
Definition:
Two triangles are congruent if there exists a sequence of rigid motions that maps one triangle onto the other.
Rigid Motions Include:
• Translation: Sliding the triangle in any direction
• Reflection: Flipping the triangle over a line
• Rotation: Turning the triangle around a point
• Combination: Any sequence of the above
NOT Rigid Motion:
• Dilation: Changing the size (creates similar, not congruent triangles)
Definition:
Two triangles are congruent if there exists a sequence of rigid motions that maps one triangle onto the other.
Rigid Motions Include:
• Translation: Sliding the triangle in any direction
• Reflection: Flipping the triangle over a line
• Rotation: Turning the triangle around a point
• Combination: Any sequence of the above
NOT Rigid Motion:
• Dilation: Changing the size (creates similar, not congruent triangles)
Steps to Prove Congruence Using Rigid Motions:
Step 1: Identify corresponding vertices in both triangles
Step 2: Describe the first rigid motion needed (usually translation to align one vertex)
Step 3: Describe additional rigid motions if needed (reflection or rotation)
Step 4: Verify that all corresponding parts now coincide
Step 5: Conclude that triangles are congruent
Step 1: Identify corresponding vertices in both triangles
Step 2: Describe the first rigid motion needed (usually translation to align one vertex)
Step 3: Describe additional rigid motions if needed (reflection or rotation)
Step 4: Verify that all corresponding parts now coincide
Step 5: Conclude that triangles are congruent
Example 1: Using translation
Given: $\triangle ABC$ and $\triangle DEF$ with all corresponding sides equal
Proof using rigid motions:
Step 1: Translate $\triangle ABC$ so that point A coincides with point D
Step 2: If needed, rotate around point D so that AB aligns with DE
Step 3: If needed, reflect over DE so that C and F are on the same side
Step 4: Since all sides are equal, point C now coincides with point F
Conclusion: $\triangle ABC \cong \triangle DEF$ (mapped by rigid motions)
Given: $\triangle ABC$ and $\triangle DEF$ with all corresponding sides equal
Proof using rigid motions:
Step 1: Translate $\triangle ABC$ so that point A coincides with point D
Step 2: If needed, rotate around point D so that AB aligns with DE
Step 3: If needed, reflect over DE so that C and F are on the same side
Step 4: Since all sides are equal, point C now coincides with point F
Conclusion: $\triangle ABC \cong \triangle DEF$ (mapped by rigid motions)
2. SSS and SAS Theorems
SSS (Side-Side-Side) Theorem
SSS Theorem: If three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent
What You Need: All three pairs of corresponding sides equal
Symbol: SSS (Side-Side-Side)
No Angles Needed: Knowing all three sides is sufficient
What You Need: All three pairs of corresponding sides equal
Symbol: SSS (Side-Side-Side)
No Angles Needed: Knowing all three sides is sufficient
SSS Congruence Theorem:
$$\text{If } \overline{AB} \cong \overline{DE}, \overline{BC} \cong \overline{EF}, \text{ and } \overline{AC} \cong \overline{DF}$$
$$\text{Then } \triangle ABC \cong \triangle DEF$$
In words: Three sides determine a unique triangle
Requirement: ALL three pairs of sides must be congruent
$$\text{If } \overline{AB} \cong \overline{DE}, \overline{BC} \cong \overline{EF}, \text{ and } \overline{AC} \cong \overline{DF}$$
$$\text{Then } \triangle ABC \cong \triangle DEF$$
In words: Three sides determine a unique triangle
Requirement: ALL three pairs of sides must be congruent
Example 1: Prove using SSS
Given:
• $\triangle ABC$ with sides AB = 5, BC = 7, AC = 9
• $\triangle PQR$ with sides PQ = 5, QR = 7, PR = 9
Proof:
AB = PQ = 5 (given)
BC = QR = 7 (given)
AC = PR = 9 (given)
All three pairs of corresponding sides are congruent.
Conclusion: $\triangle ABC \cong \triangle PQR$ by SSS
Given:
• $\triangle ABC$ with sides AB = 5, BC = 7, AC = 9
• $\triangle PQR$ with sides PQ = 5, QR = 7, PR = 9
Proof:
AB = PQ = 5 (given)
BC = QR = 7 (given)
AC = PR = 9 (given)
All three pairs of corresponding sides are congruent.
Conclusion: $\triangle ABC \cong \triangle PQR$ by SSS
SAS (Side-Angle-Side) Theorem
SAS Theorem: If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent
What You Need: Two sides AND the angle BETWEEN them
IMPORTANT: The angle MUST be between the two sides (included angle)
Symbol: SAS (Side-Angle-Side)
What You Need: Two sides AND the angle BETWEEN them
IMPORTANT: The angle MUST be between the two sides (included angle)
Symbol: SAS (Side-Angle-Side)
SAS Congruence Theorem:
$$\text{If } \overline{AB} \cong \overline{DE}, \angle B \cong \angle E, \text{ and } \overline{BC} \cong \overline{EF}$$
$$\text{Then } \triangle ABC \cong \triangle DEF$$
Key Point: The angle must be INCLUDED (between the two sides)
Pattern: Side - Angle - Side (in order around the triangle)
$$\text{If } \overline{AB} \cong \overline{DE}, \angle B \cong \angle E, \text{ and } \overline{BC} \cong \overline{EF}$$
$$\text{Then } \triangle ABC \cong \triangle DEF$$
Key Point: The angle must be INCLUDED (between the two sides)
Pattern: Side - Angle - Side (in order around the triangle)
Example 2: Prove using SAS
Given:
• In $\triangle ABC$ and $\triangle XYZ$:
• AB = 6, XY = 6
• $\angle B = 50°$, $\angle Y = 50°$
• BC = 8, YZ = 8
Proof:
AB = XY (given)
$\angle B = \angle Y$ (given)
BC = YZ (given)
Two sides and the included angle are congruent.
Conclusion: $\triangle ABC \cong \triangle XYZ$ by SAS
Given:
• In $\triangle ABC$ and $\triangle XYZ$:
• AB = 6, XY = 6
• $\angle B = 50°$, $\angle Y = 50°$
• BC = 8, YZ = 8
Proof:
AB = XY (given)
$\angle B = \angle Y$ (given)
BC = YZ (given)
Two sides and the included angle are congruent.
Conclusion: $\triangle ABC \cong \triangle XYZ$ by SAS
Common Mistake - SSA is NOT Valid!
SSA (Side-Side-Angle) does NOT prove congruence!
If you have:
• Two sides congruent
• An angle NOT between them (non-included angle)
This is called the "ambiguous case" because it can create two different triangles.
Remember: For SAS, the angle MUST be between the two sides!
SSA (Side-Side-Angle) does NOT prove congruence!
If you have:
• Two sides congruent
• An angle NOT between them (non-included angle)
This is called the "ambiguous case" because it can create two different triangles.
Remember: For SAS, the angle MUST be between the two sides!
3. Proving Triangles Congruent by SSS and SAS
Steps to Write a Congruence Proof:
Step 1: Given
List all given information
Step 2: Identify What to Prove
State which triangles you need to prove congruent
Step 3: Mark the Diagram
Mark all equal sides and angles
Step 4: Look for Shared Parts
• Reflexive sides (shared sides)
• Vertical angles
• Right angles
Step 5: Choose Criterion
Decide if you have SSS or SAS (or another criterion)
Step 6: Write Proof
List statements and reasons in logical order
Step 7: Conclusion
State the congruence with the criterion used
Step 1: Given
List all given information
Step 2: Identify What to Prove
State which triangles you need to prove congruent
Step 3: Mark the Diagram
Mark all equal sides and angles
Step 4: Look for Shared Parts
• Reflexive sides (shared sides)
• Vertical angles
• Right angles
Step 5: Choose Criterion
Decide if you have SSS or SAS (or another criterion)
Step 6: Write Proof
List statements and reasons in logical order
Step 7: Conclusion
State the congruence with the criterion used
Example 1: Two-column proof with SSS
Given: AB = DE, BC = EF, AC = DF
Prove: $\triangle ABC \cong \triangle DEF$
Given: AB = DE, BC = EF, AC = DF
Prove: $\triangle ABC \cong \triangle DEF$
| Statements | Reasons |
|---|---|
| 1. AB = DE | 1. Given |
| 2. BC = EF | 2. Given |
| 3. AC = DF | 3. Given |
| 4. $\triangle ABC \cong \triangle DEF$ | 4. SSS Congruence Theorem |
Example 2: Proof with SAS using shared side
Given: AC bisects BD at E, AB = CD, $\angle AEB = \angle CED$
Prove: $\triangle ABE \cong \triangle CDE$
Given: AC bisects BD at E, AB = CD, $\angle AEB = \angle CED$
Prove: $\triangle ABE \cong \triangle CDE$
| Statements | Reasons |
|---|---|
| 1. AB = CD | 1. Given |
| 2. AC bisects BD at E | 2. Given |
| 3. BE = ED | 3. Definition of bisector |
| 4. $\angle AEB = \angle CED$ | 4. Given (or Vertical Angles) |
| 5. $\triangle ABE \cong \triangle CDE$ | 5. SAS Congruence Theorem |
4. ASA and AAS Theorems
ASA (Angle-Side-Angle) Theorem
ASA Theorem: If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent
What You Need: Two angles AND the side BETWEEN them
IMPORTANT: The side MUST be between the two angles (included side)
Symbol: ASA (Angle-Side-Angle)
What You Need: Two angles AND the side BETWEEN them
IMPORTANT: The side MUST be between the two angles (included side)
Symbol: ASA (Angle-Side-Angle)
ASA Congruence Theorem:
$$\text{If } \angle A \cong \angle D, \overline{AB} \cong \overline{DE}, \text{ and } \angle B \cong \angle E$$
$$\text{Then } \triangle ABC \cong \triangle DEF$$
Key Point: The side must be INCLUDED (between the two angles)
Pattern: Angle - Side - Angle (in order around the triangle)
$$\text{If } \angle A \cong \angle D, \overline{AB} \cong \overline{DE}, \text{ and } \angle B \cong \angle E$$
$$\text{Then } \triangle ABC \cong \triangle DEF$$
Key Point: The side must be INCLUDED (between the two angles)
Pattern: Angle - Side - Angle (in order around the triangle)
Example 1: Prove using ASA
Given:
• $\angle A = 40°$, $\angle D = 40°$
• AB = 10, DE = 10
• $\angle B = 60°$, $\angle E = 60°$
Proof:
$\angle A = \angle D$ (given)
AB = DE (given)
$\angle B = \angle E$ (given)
Two angles and the included side are congruent.
Conclusion: $\triangle ABC \cong \triangle DEF$ by ASA
Given:
• $\angle A = 40°$, $\angle D = 40°$
• AB = 10, DE = 10
• $\angle B = 60°$, $\angle E = 60°$
Proof:
$\angle A = \angle D$ (given)
AB = DE (given)
$\angle B = \angle E$ (given)
Two angles and the included side are congruent.
Conclusion: $\triangle ABC \cong \triangle DEF$ by ASA
AAS (Angle-Angle-Side) Theorem
AAS Theorem: If two angles and a non-included side of one triangle are congruent to two angles and a non-included side of another triangle, then the triangles are congruent
What You Need: Two angles AND a side NOT between them
Difference from ASA: The side is NOT between the two angles
Symbol: AAS (Angle-Angle-Side)
What You Need: Two angles AND a side NOT between them
Difference from ASA: The side is NOT between the two angles
Symbol: AAS (Angle-Angle-Side)
AAS Congruence Theorem:
$$\text{If } \angle A \cong \angle D, \angle B \cong \angle E, \text{ and } \overline{BC} \cong \overline{EF}$$
$$\text{Then } \triangle ABC \cong \triangle DEF$$
Key Point: The side is NOT between the two angles
Why It Works: If two angles are equal, the third angle must also be equal (Angle Sum Theorem), so you really have ASA
$$\text{If } \angle A \cong \angle D, \angle B \cong \angle E, \text{ and } \overline{BC} \cong \overline{EF}$$
$$\text{Then } \triangle ABC \cong \triangle DEF$$
Key Point: The side is NOT between the two angles
Why It Works: If two angles are equal, the third angle must also be equal (Angle Sum Theorem), so you really have ASA
Example 2: Prove using AAS
Given:
• $\angle P = 50°$, $\angle X = 50°$
• $\angle Q = 70°$, $\angle Y = 70°$
• QR = 8, YZ = 8 (side NOT between the angles)
Proof:
$\angle P = \angle X$ (given)
$\angle Q = \angle Y$ (given)
QR = YZ (given, non-included side)
Two angles and a non-included side are congruent.
Conclusion: $\triangle PQR \cong \triangle XYZ$ by AAS
Given:
• $\angle P = 50°$, $\angle X = 50°$
• $\angle Q = 70°$, $\angle Y = 70°$
• QR = 8, YZ = 8 (side NOT between the angles)
Proof:
$\angle P = \angle X$ (given)
$\angle Q = \angle Y$ (given)
QR = YZ (given, non-included side)
Two angles and a non-included side are congruent.
Conclusion: $\triangle PQR \cong \triangle XYZ$ by AAS
Understanding the Difference: ASA vs. AAS
ASA: Angle - Side - Angle
The side is BETWEEN the two angles
Pattern: ∠-side-∠
AAS: Angle - Angle - Side
The side is NOT between the two angles
Pattern: ∠-∠-side
Both are valid! Both prove triangle congruence.
ASA: Angle - Side - Angle
The side is BETWEEN the two angles
Pattern: ∠-side-∠
AAS: Angle - Angle - Side
The side is NOT between the two angles
Pattern: ∠-∠-side
Both are valid! Both prove triangle congruence.
5. Proving Triangles Congruent by ASA and AAS
Example 1: Two-column proof with ASA
Given: $\angle A = \angle D$, AB = DE, $\angle B = \angle E$
Prove: $\triangle ABC \cong \triangle DEF$
Given: $\angle A = \angle D$, AB = DE, $\angle B = \angle E$
Prove: $\triangle ABC \cong \triangle DEF$
| Statements | Reasons |
|---|---|
| 1. $\angle A = \angle D$ | 1. Given |
| 2. AB = DE | 2. Given |
| 3. $\angle B = \angle E$ | 3. Given |
| 4. $\triangle ABC \cong \triangle DEF$ | 4. ASA Congruence Theorem |
Example 2: Proof with AAS using vertical angles
Given: Lines AB and CD intersect at E, AE = CE, $\angle A = \angle C$
Prove: $\triangle AEB \cong \triangle CED$
Given: Lines AB and CD intersect at E, AE = CE, $\angle A = \angle C$
Prove: $\triangle AEB \cong \triangle CED$
| Statements | Reasons |
|---|---|
| 1. $\angle A = \angle C$ | 1. Given |
| 2. $\angle AEB = \angle CED$ | 2. Vertical Angles are congruent |
| 3. AE = CE | 3. Given |
| 4. $\triangle AEB \cong \triangle CED$ | 4. AAS Congruence Theorem |
6. SSS, SAS, ASA, and AAS Theorems - Summary
Four Main Triangle Congruence Theorems:
1. SSS (Side-Side-Side):
All three pairs of corresponding sides are congruent
$$\overline{AB} \cong \overline{DE}, \overline{BC} \cong \overline{EF}, \overline{AC} \cong \overline{DF}$$
2. SAS (Side-Angle-Side):
Two pairs of sides and the INCLUDED angle are congruent
$$\overline{AB} \cong \overline{DE}, \angle B \cong \angle E, \overline{BC} \cong \overline{EF}$$
3. ASA (Angle-Side-Angle):
Two pairs of angles and the INCLUDED side are congruent
$$\angle A \cong \angle D, \overline{AB} \cong \overline{DE}, \angle B \cong \angle E$$
4. AAS (Angle-Angle-Side):
Two pairs of angles and a NON-INCLUDED side are congruent
$$\angle A \cong \angle D, \angle B \cong \angle E, \overline{BC} \cong \overline{EF}$$
1. SSS (Side-Side-Side):
All three pairs of corresponding sides are congruent
$$\overline{AB} \cong \overline{DE}, \overline{BC} \cong \overline{EF}, \overline{AC} \cong \overline{DF}$$
2. SAS (Side-Angle-Side):
Two pairs of sides and the INCLUDED angle are congruent
$$\overline{AB} \cong \overline{DE}, \angle B \cong \angle E, \overline{BC} \cong \overline{EF}$$
3. ASA (Angle-Side-Angle):
Two pairs of angles and the INCLUDED side are congruent
$$\angle A \cong \angle D, \overline{AB} \cong \overline{DE}, \angle B \cong \angle E$$
4. AAS (Angle-Angle-Side):
Two pairs of angles and a NON-INCLUDED side are congruent
$$\angle A \cong \angle D, \angle B \cong \angle E, \overline{BC} \cong \overline{EF}$$
What Does NOT Work:
• AAA (Angle-Angle-Angle): Creates similar triangles, not necessarily congruent
• SSA (Side-Side-Angle): Ambiguous - can create two different triangles
• ASS: Same as SSA - does NOT work
Remember: You need at least ONE side to prove congruence!
• AAA (Angle-Angle-Angle): Creates similar triangles, not necessarily congruent
• SSA (Side-Side-Angle): Ambiguous - can create two different triangles
• ASS: Same as SSA - does NOT work
Remember: You need at least ONE side to prove congruence!
7. SSS Theorem in the Coordinate Plane
SSS in Coordinate Plane: Using distance formula to find side lengths, then comparing
Distance Formula: Used to calculate length of each side
Goal: Prove all three pairs of sides are equal
Distance Formula: Used to calculate length of each side
Goal: Prove all three pairs of sides are equal
Distance Formula:
For two points $(x_1, y_1)$ and $(x_2, y_2)$:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Steps to Prove SSS in Coordinate Plane:
1. Find the length of all three sides of $\triangle ABC$ using distance formula
2. Find the length of all three sides of $\triangle DEF$ using distance formula
3. Compare corresponding sides
4. If all three pairs match, triangles are congruent by SSS
For two points $(x_1, y_1)$ and $(x_2, y_2)$:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Steps to Prove SSS in Coordinate Plane:
1. Find the length of all three sides of $\triangle ABC$ using distance formula
2. Find the length of all three sides of $\triangle DEF$ using distance formula
3. Compare corresponding sides
4. If all three pairs match, triangles are congruent by SSS
Example: SSS in coordinate plane
Given:
• $\triangle ABC$ with A(0, 0), B(3, 0), C(0, 4)
• $\triangle DEF$ with D(5, 1), E(8, 1), F(5, 5)
Prove: $\triangle ABC \cong \triangle DEF$
Solution:
For $\triangle ABC$:
$AB = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{9} = 3$
$BC = \sqrt{(0-3)^2 + (4-0)^2} = \sqrt{9+16} = \sqrt{25} = 5$
$AC = \sqrt{(0-0)^2 + (4-0)^2} = \sqrt{16} = 4$
For $\triangle DEF$:
$DE = \sqrt{(8-5)^2 + (1-1)^2} = \sqrt{9} = 3$
$EF = \sqrt{(5-8)^2 + (5-1)^2} = \sqrt{9+16} = \sqrt{25} = 5$
$DF = \sqrt{(5-5)^2 + (5-1)^2} = \sqrt{16} = 4$
Comparison:
AB = DE = 3
BC = EF = 5
AC = DF = 4
Conclusion: $\triangle ABC \cong \triangle DEF$ by SSS
Given:
• $\triangle ABC$ with A(0, 0), B(3, 0), C(0, 4)
• $\triangle DEF$ with D(5, 1), E(8, 1), F(5, 5)
Prove: $\triangle ABC \cong \triangle DEF$
Solution:
For $\triangle ABC$:
$AB = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{9} = 3$
$BC = \sqrt{(0-3)^2 + (4-0)^2} = \sqrt{9+16} = \sqrt{25} = 5$
$AC = \sqrt{(0-0)^2 + (4-0)^2} = \sqrt{16} = 4$
For $\triangle DEF$:
$DE = \sqrt{(8-5)^2 + (1-1)^2} = \sqrt{9} = 3$
$EF = \sqrt{(5-8)^2 + (5-1)^2} = \sqrt{9+16} = \sqrt{25} = 5$
$DF = \sqrt{(5-5)^2 + (5-1)^2} = \sqrt{16} = 4$
Comparison:
AB = DE = 3
BC = EF = 5
AC = DF = 4
Conclusion: $\triangle ABC \cong \triangle DEF$ by SSS
8. Proofs Involving Corresponding Parts of Congruent Triangles (CPCTC)
CPCTC: "Corresponding Parts of Congruent Triangles are Congruent"
When to Use: AFTER proving triangles are congruent
Purpose: To prove that specific sides or angles are equal
Important: Must prove triangle congruence FIRST, then use CPCTC
When to Use: AFTER proving triangles are congruent
Purpose: To prove that specific sides or angles are equal
Important: Must prove triangle congruence FIRST, then use CPCTC
Two-Step Process for CPCTC Proofs:
Step 1: Prove Triangle Congruence
Use one of the congruence theorems (SSS, SAS, ASA, AAS, or HL) to prove the triangles are congruent
Step 2: Apply CPCTC
Once triangles are proven congruent, state that the desired parts are congruent by CPCTC
Common Uses:
• Proving two segments are equal
• Proving two angles are equal
• Proving lines are parallel (using congruent alternate interior angles)
• Proving lines are perpendicular (using congruent right angles)
Step 1: Prove Triangle Congruence
Use one of the congruence theorems (SSS, SAS, ASA, AAS, or HL) to prove the triangles are congruent
Step 2: Apply CPCTC
Once triangles are proven congruent, state that the desired parts are congruent by CPCTC
Common Uses:
• Proving two segments are equal
• Proving two angles are equal
• Proving lines are parallel (using congruent alternate interior angles)
• Proving lines are perpendicular (using congruent right angles)
Example: CPCTC proof
Given: AB = CD, BC = DA
Prove: $\angle A = \angle C$
Given: AB = CD, BC = DA
Prove: $\angle A = \angle C$
| Statements | Reasons |
|---|---|
| 1. AB = CD | 1. Given |
| 2. BC = DA | 2. Given |
| 3. AC = AC | 3. Reflexive Property |
| 4. $\triangle ABC \cong \triangle CDA$ | 4. SSS Congruence Theorem |
| 5. $\angle A = \angle C$ | 5. CPCTC |
9. Congruency in Isosceles and Equilateral Triangles
Isosceles Triangle Properties
Isosceles Triangle: A triangle with two equal sides
Legs: The two equal sides
Base: The third side
Vertex Angle: The angle between the two equal sides
Base Angles: The two angles opposite the equal sides
Legs: The two equal sides
Base: The third side
Vertex Angle: The angle between the two equal sides
Base Angles: The two angles opposite the equal sides
Isosceles Triangle Theorem:
$$\text{If two sides of a triangle are congruent, then the angles opposite those sides are congruent.}$$
In other words:
If AB = AC, then $\angle B = \angle C$
Converse of Isosceles Triangle Theorem:
$$\text{If two angles of a triangle are congruent, then the sides opposite those angles are congruent.}$$
In other words:
If $\angle B = \angle C$, then AB = AC
$$\text{If two sides of a triangle are congruent, then the angles opposite those sides are congruent.}$$
In other words:
If AB = AC, then $\angle B = \angle C$
Converse of Isosceles Triangle Theorem:
$$\text{If two angles of a triangle are congruent, then the sides opposite those angles are congruent.}$$
In other words:
If $\angle B = \angle C$, then AB = AC
Example 1: Find missing angle in isosceles triangle
Given: Isosceles $\triangle ABC$ with AB = AC, $\angle A = 40°$
Find: $\angle B$ and $\angle C$
Solution:
Since AB = AC, the base angles are equal:
$\angle B = \angle C$
Using Angle Sum Theorem:
$40° + \angle B + \angle C = 180°$
$40° + 2\angle B = 180°$
$2\angle B = 140°$
$\angle B = 70°$
Answer: $\angle B = \angle C = 70°$
Given: Isosceles $\triangle ABC$ with AB = AC, $\angle A = 40°$
Find: $\angle B$ and $\angle C$
Solution:
Since AB = AC, the base angles are equal:
$\angle B = \angle C$
Using Angle Sum Theorem:
$40° + \angle B + \angle C = 180°$
$40° + 2\angle B = 180°$
$2\angle B = 140°$
$\angle B = 70°$
Answer: $\angle B = \angle C = 70°$
Equilateral Triangle Properties
Equilateral Triangle: A triangle with all three sides equal
Special Property: All three angles are also equal (60° each)
Also Equiangular: Equal angles means it's equiangular
Highly Symmetric: Has 3 lines of symmetry
Special Property: All three angles are also equal (60° each)
Also Equiangular: Equal angles means it's equiangular
Highly Symmetric: Has 3 lines of symmetry
Equilateral Triangle Theorem:
$$\text{If a triangle is equilateral, then it is equiangular.}$$
All sides equal → All angles equal
If AB = BC = AC, then $\angle A = \angle B = \angle C = 60°$
Converse:
$$\text{If a triangle is equiangular, then it is equilateral.}$$
All angles equal → All sides equal
If $\angle A = \angle B = \angle C$, then AB = BC = AC
$$\text{If a triangle is equilateral, then it is equiangular.}$$
All sides equal → All angles equal
If AB = BC = AC, then $\angle A = \angle B = \angle C = 60°$
Converse:
$$\text{If a triangle is equiangular, then it is equilateral.}$$
All angles equal → All sides equal
If $\angle A = \angle B = \angle C$, then AB = BC = AC
10. Proofs Involving Isosceles Triangles
Example 1: Prove base angles are equal
Given: $\triangle ABC$ with AB = AC, D is midpoint of BC
Prove: $\angle B = \angle C$
Given: $\triangle ABC$ with AB = AC, D is midpoint of BC
Prove: $\angle B = \angle C$
| Statements | Reasons |
|---|---|
| 1. AB = AC | 1. Given |
| 2. D is midpoint of BC | 2. Given |
| 3. BD = DC | 3. Definition of midpoint |
| 4. AD = AD | 4. Reflexive Property |
| 5. $\triangle ABD \cong \triangle ACD$ | 5. SSS Congruence Theorem |
| 6. $\angle B = \angle C$ | 6. CPCTC |
Example 2: Using isosceles triangle properties
Given: $\triangle PQR$ with PQ = PR, $\angle Q = 3x + 10$, $\angle R = 5x - 20$
Find: x and the measures of $\angle Q$ and $\angle R$
Solution:
Since PQ = PR, base angles are equal:
$\angle Q = \angle R$
$3x + 10 = 5x - 20$
$10 + 20 = 5x - 3x$
$30 = 2x$
$x = 15$
$\angle Q = 3(15) + 10 = 55°$
$\angle R = 5(15) - 20 = 55°$ ✓
Answer: x = 15, $\angle Q = \angle R = 55°$
Given: $\triangle PQR$ with PQ = PR, $\angle Q = 3x + 10$, $\angle R = 5x - 20$
Find: x and the measures of $\angle Q$ and $\angle R$
Solution:
Since PQ = PR, base angles are equal:
$\angle Q = \angle R$
$3x + 10 = 5x - 20$
$10 + 20 = 5x - 3x$
$30 = 2x$
$x = 15$
$\angle Q = 3(15) + 10 = 55°$
$\angle R = 5(15) - 20 = 55°$ ✓
Answer: x = 15, $\angle Q = \angle R = 55°$
11. Hypotenuse-Leg (HL) Theorem
HL Theorem: A congruence theorem for RIGHT TRIANGLES ONLY
What You Need: Hypotenuse and one leg of each right triangle
Applies to: RIGHT triangles only (one 90° angle)
Also Called: RHS (Right angle-Hypotenuse-Side) in some countries
What You Need: Hypotenuse and one leg of each right triangle
Applies to: RIGHT triangles only (one 90° angle)
Also Called: RHS (Right angle-Hypotenuse-Side) in some countries
Hypotenuse-Leg (HL) Theorem:
$$\text{If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse}$$
$$\text{and a leg of another right triangle, then the triangles are congruent.}$$
Requirements:
• Both triangles must be RIGHT triangles
• Hypotenuses must be congruent
• One pair of legs must be congruent
In symbols:
If $\angle B = \angle E = 90°$, AC = DF (hypotenuses), and AB = DE (legs)
Then $\triangle ABC \cong \triangle DEF$
$$\text{If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse}$$
$$\text{and a leg of another right triangle, then the triangles are congruent.}$$
Requirements:
• Both triangles must be RIGHT triangles
• Hypotenuses must be congruent
• One pair of legs must be congruent
In symbols:
If $\angle B = \angle E = 90°$, AC = DF (hypotenuses), and AB = DE (legs)
Then $\triangle ABC \cong \triangle DEF$
Why HL Works:
The HL theorem is actually a special case of SSS:
1. You know the hypotenuse and one leg
2. Using the Pythagorean Theorem, you can calculate the other leg:
$$a^2 + b^2 = c^2$$
3. Since both triangles have the same hypotenuse and same leg, they must have the same other leg
4. Therefore, all three sides are equal (SSS)
HL is a shortcut that only works for right triangles!
The HL theorem is actually a special case of SSS:
1. You know the hypotenuse and one leg
2. Using the Pythagorean Theorem, you can calculate the other leg:
$$a^2 + b^2 = c^2$$
3. Since both triangles have the same hypotenuse and same leg, they must have the same other leg
4. Therefore, all three sides are equal (SSS)
HL is a shortcut that only works for right triangles!
Example 1: Prove using HL
Given:
• $\triangle ABC$ and $\triangle DEF$ are right triangles
• $\angle B = \angle E = 90°$
• AC = 10, DF = 10 (hypotenuses)
• AB = 6, DE = 6 (legs)
Proof:
Both triangles are right triangles (given)
AC = DF = 10 (hypotenuses are equal)
AB = DE = 6 (one pair of legs is equal)
Conclusion: $\triangle ABC \cong \triangle DEF$ by HL Theorem
Given:
• $\triangle ABC$ and $\triangle DEF$ are right triangles
• $\angle B = \angle E = 90°$
• AC = 10, DF = 10 (hypotenuses)
• AB = 6, DE = 6 (legs)
Proof:
Both triangles are right triangles (given)
AC = DF = 10 (hypotenuses are equal)
AB = DE = 6 (one pair of legs is equal)
Conclusion: $\triangle ABC \cong \triangle DEF$ by HL Theorem
Example 2: Two-column proof with HL
Given: $\angle B$ and $\angle D$ are right angles, AC = CE, BC = DC
Prove: $\triangle ABC \cong \triangle EDC$
Given: $\angle B$ and $\angle D$ are right angles, AC = CE, BC = DC
Prove: $\triangle ABC \cong \triangle EDC$
| Statements | Reasons |
|---|---|
| 1. $\angle B$ and $\angle D$ are right angles | 1. Given |
| 2. $\triangle ABC$ and $\triangle EDC$ are right triangles | 2. Definition of right triangle |
| 3. AC = CE | 3. Given (hypotenuses) |
| 4. BC = DC | 4. Given (legs) |
| 5. $\triangle ABC \cong \triangle EDC$ | 5. HL Theorem |
Remember for HL:
• Works ONLY for right triangles
• Need hypotenuse + ONE leg (not both legs - that would be SAS)
• The right angle doesn't need to be stated if it's marked in the diagram
• Very useful for proofs involving perpendicular lines or right angles
• Works ONLY for right triangles
• Need hypotenuse + ONE leg (not both legs - that would be SAS)
• The right angle doesn't need to be stated if it's marked in the diagram
• Very useful for proofs involving perpendicular lines or right angles
Triangle Congruence Theorems Summary
| Theorem | What's Needed | Diagram Marks | Special Notes |
|---|---|---|---|
| SSS | 3 sides | All sides marked equal | No angles needed |
| SAS | 2 sides + INCLUDED angle | 2 sides and angle between them | Angle MUST be included |
| ASA | 2 angles + INCLUDED side | 2 angles and side between them | Side MUST be included |
| AAS | 2 angles + NON-INCLUDED side | 2 angles and any side | Side NOT between angles |
| HL | Hypotenuse + 1 leg (right triangles) | Right angle + hypotenuse + leg | RIGHT TRIANGLES ONLY |
What Does NOT Prove Congruence
| Combination | Why It Doesn't Work | What It Does Prove |
|---|---|---|
| AAA | Doesn't determine size | SIMILARITY (not congruence) |
| SSA | Ambiguous case - can make 2 different triangles | Nothing (not valid) |
| ASS | Same as SSA | Nothing (not valid) |
Common Proof Strategies
| Strategy | When to Use | Example |
|---|---|---|
| Reflexive Property | Triangles share a common side | AC = AC (same segment in both triangles) |
| Vertical Angles | Two lines intersect forming triangles | ∠AEB = ∠CED (vertical angles) |
| Midpoint | Point divides segment in half | If M is midpoint, AM = MB |
| Bisector | Line/ray divides angle in half | If AD bisects ∠A, ∠1 = ∠2 |
| Isosceles Triangle | Triangle has two equal sides | If AB = AC, then ∠B = ∠C |
| Right Angle | Perpendicular lines | If AB ⊥ CD, then ∠ABC = 90° |
Isosceles and Equilateral Properties
| Triangle Type | Definition | Properties | Angle Measures |
|---|---|---|---|
| Isosceles | 2 equal sides | Base angles are equal | Two angles equal, third different |
| Equilateral | 3 equal sides | All angles equal (equiangular) | All angles = 60° |
| Right Isosceles | 2 equal sides + 90° angle | 45°-45°-90° triangle | Two 45° angles, one 90° angle |
Key Formulas Quick Reference
| Formula/Theorem | Statement | Use |
|---|---|---|
| Distance Formula | $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ | Find side lengths in coordinate plane |
| CPCTC | Corresponding Parts of Congruent Triangles are Congruent | After proving congruence, conclude parts equal |
| Pythagorean Theorem | $a^2 + b^2 = c^2$ | Find missing side in right triangles |
| Triangle Angle Sum | Sum of angles = 180° | Find missing angles |
| Isosceles Triangle | If 2 sides equal, opposite angles equal | Find equal angles |
| Equilateral Triangle | All sides equal → all angles = 60° | Identify equiangular property |
Success Tips for Triangle Congruence:
✓ Five valid methods: SSS, SAS, ASA, AAS, HL (right triangles only)
✓ AAA and SSA do NOT work for congruence
✓ SAS: angle must be BETWEEN the two sides (included angle)
✓ ASA: side must be BETWEEN the two angles (included side)
✓ HL: Only for RIGHT triangles (hypotenuse + one leg)
✓ Use CPCTC AFTER proving triangle congruence
✓ Rigid motions (translation, reflection, rotation) prove congruence
✓ Isosceles triangle: equal sides → equal base angles
✓ Equilateral triangle: all sides equal → all angles 60°
✓ Look for reflexive sides, vertical angles, and midpoints in proofs!
✓ Five valid methods: SSS, SAS, ASA, AAS, HL (right triangles only)
✓ AAA and SSA do NOT work for congruence
✓ SAS: angle must be BETWEEN the two sides (included angle)
✓ ASA: side must be BETWEEN the two angles (included side)
✓ HL: Only for RIGHT triangles (hypotenuse + one leg)
✓ Use CPCTC AFTER proving triangle congruence
✓ Rigid motions (translation, reflection, rotation) prove congruence
✓ Isosceles triangle: equal sides → equal base angles
✓ Equilateral triangle: all sides equal → all angles 60°
✓ Look for reflexive sides, vertical angles, and midpoints in proofs!