Triangles Review - Tenth Grade Geometry
Introduction to Triangles
Triangle: A polygon with three sides, three vertices, and three angles
Symbol: △ABC
Vertices: The corner points (A, B, C)
Sides: The line segments connecting vertices
Interior Angles: The three angles inside the triangle
Sum of Angles: Always equals 180°
Symbol: △ABC
Vertices: The corner points (A, B, C)
Sides: The line segments connecting vertices
Interior Angles: The three angles inside the triangle
Sum of Angles: Always equals 180°
1. Classify Triangles
Classification Methods:
1. By Sides: Based on length of sides
2. By Angles: Based on measure of angles
3. Combined: Both sides and angles
Key Concept: A triangle can be classified in multiple ways simultaneously
1. By Sides: Based on length of sides
2. By Angles: Based on measure of angles
3. Combined: Both sides and angles
Key Concept: A triangle can be classified in multiple ways simultaneously
Classification by Sides
1. Equilateral Triangle:
• All three sides are equal in length
• Symbol: $AB = BC = CA$
• All three angles are equal (each = 60°)
• Most symmetrical triangle
• Also called "regular triangle"
2. Isosceles Triangle:
• Exactly two sides are equal in length
• The two equal sides are called legs
• The third side is called the base
• The two angles opposite the equal sides are equal
• These equal angles are called base angles
• The angle between the two equal sides is called the vertex angle or apex angle
3. Scalene Triangle:
• All three sides have different lengths
• Symbol: $AB \neq BC \neq CA$
• All three angles have different measures
• No line of symmetry
• Most general type of triangle
• All three sides are equal in length
• Symbol: $AB = BC = CA$
• All three angles are equal (each = 60°)
• Most symmetrical triangle
• Also called "regular triangle"
2. Isosceles Triangle:
• Exactly two sides are equal in length
• The two equal sides are called legs
• The third side is called the base
• The two angles opposite the equal sides are equal
• These equal angles are called base angles
• The angle between the two equal sides is called the vertex angle or apex angle
3. Scalene Triangle:
• All three sides have different lengths
• Symbol: $AB \neq BC \neq CA$
• All three angles have different measures
• No line of symmetry
• Most general type of triangle
Example 1: Classify by sides
Triangle with sides: 5 cm, 5 cm, 5 cm
Classification: Equilateral (all sides equal)
Triangle with sides: 7 cm, 7 cm, 10 cm
Classification: Isosceles (two sides equal)
Triangle with sides: 3 cm, 4 cm, 5 cm
Classification: Scalene (all sides different)
Triangle with sides: 5 cm, 5 cm, 5 cm
Classification: Equilateral (all sides equal)
Triangle with sides: 7 cm, 7 cm, 10 cm
Classification: Isosceles (two sides equal)
Triangle with sides: 3 cm, 4 cm, 5 cm
Classification: Scalene (all sides different)
Classification by Angles
1. Acute Triangle:
• All three angles are less than 90°
• Each angle: $0° < \theta < 90°$
• All angles are acute
• Example: Triangle with angles 60°, 70°, 50°
2. Right Triangle:
• Exactly one angle equals 90°
• The side opposite the right angle is called the hypotenuse (longest side)
• The other two sides are called legs
• Follows Pythagorean Theorem: $a^2 + b^2 = c^2$
• The two acute angles are complementary (sum = 90°)
• Symbol: Small square at right angle
3. Obtuse Triangle:
• Exactly one angle is greater than 90°
• Obtuse angle: $90° < \theta < 180°$
• The other two angles must be acute
• Example: Triangle with angles 110°, 40°, 30°
4. Equiangular Triangle:
• All three angles are equal (each = 60°)
• Also an equilateral triangle
• Special case of acute triangle
• All three angles are less than 90°
• Each angle: $0° < \theta < 90°$
• All angles are acute
• Example: Triangle with angles 60°, 70°, 50°
2. Right Triangle:
• Exactly one angle equals 90°
• The side opposite the right angle is called the hypotenuse (longest side)
• The other two sides are called legs
• Follows Pythagorean Theorem: $a^2 + b^2 = c^2$
• The two acute angles are complementary (sum = 90°)
• Symbol: Small square at right angle
3. Obtuse Triangle:
• Exactly one angle is greater than 90°
• Obtuse angle: $90° < \theta < 180°$
• The other two angles must be acute
• Example: Triangle with angles 110°, 40°, 30°
4. Equiangular Triangle:
• All three angles are equal (each = 60°)
• Also an equilateral triangle
• Special case of acute triangle
Example 2: Classify by angles
Triangle with angles: 45°, 55°, 80°
All angles < 90° → Acute Triangle
Triangle with angles: 90°, 60°, 30°
One angle = 90° → Right Triangle
Triangle with angles: 120°, 35°, 25°
One angle > 90° → Obtuse Triangle
Triangle with angles: 60°, 60°, 60°
All angles equal → Equiangular Triangle
Triangle with angles: 45°, 55°, 80°
All angles < 90° → Acute Triangle
Triangle with angles: 90°, 60°, 30°
One angle = 90° → Right Triangle
Triangle with angles: 120°, 35°, 25°
One angle > 90° → Obtuse Triangle
Triangle with angles: 60°, 60°, 60°
All angles equal → Equiangular Triangle
Combined Classification
Triangles can be classified using BOTH sides and angles:
• Acute Scalene Triangle: All different sides, all acute angles
• Acute Isosceles Triangle: Two equal sides, all acute angles
• Right Scalene Triangle: All different sides, one right angle
• Right Isosceles Triangle: Two equal sides (legs), one right angle (45°-45°-90°)
• Obtuse Scalene Triangle: All different sides, one obtuse angle
• Obtuse Isosceles Triangle: Two equal sides, one obtuse angle
• Equilateral Triangle: All equal sides, all equal angles (automatically equiangular and acute)
• Acute Scalene Triangle: All different sides, all acute angles
• Acute Isosceles Triangle: Two equal sides, all acute angles
• Right Scalene Triangle: All different sides, one right angle
• Right Isosceles Triangle: Two equal sides (legs), one right angle (45°-45°-90°)
• Obtuse Scalene Triangle: All different sides, one obtuse angle
• Obtuse Isosceles Triangle: Two equal sides, one obtuse angle
• Equilateral Triangle: All equal sides, all equal angles (automatically equiangular and acute)
Example 3: Combined classification
Triangle with sides 5, 5, 7 and angles 45°, 45°, 90°
• By sides: Isosceles (two sides equal)
• By angles: Right (one 90° angle)
• Classification: Right Isosceles Triangle
Triangle with all sides equal and all angles 60°
• By sides: Equilateral
• By angles: Equiangular (also acute)
• Classification: Equilateral Triangle
Triangle with sides 5, 5, 7 and angles 45°, 45°, 90°
• By sides: Isosceles (two sides equal)
• By angles: Right (one 90° angle)
• Classification: Right Isosceles Triangle
Triangle with all sides equal and all angles 60°
• By sides: Equilateral
• By angles: Equiangular (also acute)
• Classification: Equilateral Triangle
2. Triangle Angle-Sum Theorem
Triangle Angle-Sum Theorem: The most fundamental property of triangles
Also called: Triangle Sum Theorem or Angle Sum Property
Application: Finding unknown angles in triangles
Also called: Triangle Sum Theorem or Angle Sum Property
Application: Finding unknown angles in triangles
Triangle Angle-Sum Theorem:
Statement: The sum of the measures of the interior angles of a triangle is always 180°.
For any triangle ABC:
$$\angle A + \angle B + \angle C = 180°$$
Or using measure notation:
$$m\angle A + m\angle B + m\angle C = 180°$$
This is true for ALL triangles: acute, right, obtuse, scalene, isosceles, equilateral
Statement: The sum of the measures of the interior angles of a triangle is always 180°.
For any triangle ABC:
$$\angle A + \angle B + \angle C = 180°$$
Or using measure notation:
$$m\angle A + m\angle B + m\angle C = 180°$$
This is true for ALL triangles: acute, right, obtuse, scalene, isosceles, equilateral
Steps to Find Unknown Angle:
Step 1: Identify the given angles
Step 2: Write equation: Sum of all three angles = 180°
Step 3: Substitute known values
Step 4: Solve for unknown angle
Step 5: Check that answer makes sense (should be between 0° and 180°)
Step 1: Identify the given angles
Step 2: Write equation: Sum of all three angles = 180°
Step 3: Substitute known values
Step 4: Solve for unknown angle
Step 5: Check that answer makes sense (should be between 0° and 180°)
Example 1: Find missing angle
In △ABC, if ∠A = 50° and ∠B = 70°, find ∠C
Using Triangle Angle-Sum Theorem:
$\angle A + \angle B + \angle C = 180°$
$50° + 70° + \angle C = 180°$
$120° + \angle C = 180°$
$\angle C = 180° - 120°$
$\angle C = 60°$
Answer: ∠C = 60°
Check: 50° + 70° + 60° = 180° ✓
In △ABC, if ∠A = 50° and ∠B = 70°, find ∠C
Using Triangle Angle-Sum Theorem:
$\angle A + \angle B + \angle C = 180°$
$50° + 70° + \angle C = 180°$
$120° + \angle C = 180°$
$\angle C = 180° - 120°$
$\angle C = 60°$
Answer: ∠C = 60°
Check: 50° + 70° + 60° = 180° ✓
Example 2: Using algebra
In △PQR, if ∠P = 2x, ∠Q = 3x, and ∠R = 4x, find x and all angles
$\angle P + \angle Q + \angle R = 180°$
$2x + 3x + 4x = 180°$
$9x = 180°$
$x = 20°$
Find each angle:
$\angle P = 2(20°) = 40°$
$\angle Q = 3(20°) = 60°$
$\angle R = 4(20°) = 80°$
Check: 40° + 60° + 80° = 180° ✓
In △PQR, if ∠P = 2x, ∠Q = 3x, and ∠R = 4x, find x and all angles
$\angle P + \angle Q + \angle R = 180°$
$2x + 3x + 4x = 180°$
$9x = 180°$
$x = 20°$
Find each angle:
$\angle P = 2(20°) = 40°$
$\angle Q = 3(20°) = 60°$
$\angle R = 4(20°) = 80°$
Check: 40° + 60° + 80° = 180° ✓
Example 3: Right triangle
In a right triangle, one angle is 35°. Find the third angle.
Given: One angle = 90° (right angle), another = 35°
$90° + 35° + x = 180°$
$125° + x = 180°$
$x = 55°$
Answer: Third angle = 55°
Note: In a right triangle, the two acute angles are complementary (sum = 90°)
In a right triangle, one angle is 35°. Find the third angle.
Given: One angle = 90° (right angle), another = 35°
$90° + 35° + x = 180°$
$125° + x = 180°$
$x = 55°$
Answer: Third angle = 55°
Note: In a right triangle, the two acute angles are complementary (sum = 90°)
Example 4: Isosceles triangle
In an isosceles triangle, the vertex angle is 40°. Find the base angles.
In isosceles triangle, base angles are equal.
Let each base angle = $x$
$40° + x + x = 180°$
$40° + 2x = 180°$
$2x = 140°$
$x = 70°$
Answer: Each base angle = 70°
In an isosceles triangle, the vertex angle is 40°. Find the base angles.
In isosceles triangle, base angles are equal.
Let each base angle = $x$
$40° + x + x = 180°$
$40° + 2x = 180°$
$2x = 140°$
$x = 70°$
Answer: Each base angle = 70°
3. Exterior Angle Theorem
Exterior Angle: Angle formed between one side of triangle and extension of adjacent side
Remote Interior Angles: The two interior angles NOT adjacent to the exterior angle
Adjacent Interior Angle: The interior angle next to the exterior angle
Key Property: Exterior and adjacent interior angles are supplementary
Remote Interior Angles: The two interior angles NOT adjacent to the exterior angle
Adjacent Interior Angle: The interior angle next to the exterior angle
Key Property: Exterior and adjacent interior angles are supplementary
Exterior Angle Theorem:
Statement: The measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote (non-adjacent) interior angles.
For triangle ABC with exterior angle at C:
$$\text{Exterior Angle} = \angle A + \angle B$$
Or more generally:
$$\angle \text{ext} = \angle \text{remote 1} + \angle \text{remote 2}$$
Related Property:
Exterior angle + Adjacent interior angle = 180° (linear pair)
Statement: The measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote (non-adjacent) interior angles.
For triangle ABC with exterior angle at C:
$$\text{Exterior Angle} = \angle A + \angle B$$
Or more generally:
$$\angle \text{ext} = \angle \text{remote 1} + \angle \text{remote 2}$$
Related Property:
Exterior angle + Adjacent interior angle = 180° (linear pair)
Key Facts About Exterior Angles:
1. Each triangle has 6 exterior angles (two at each vertex)
2. Only one exterior angle at each vertex is typically considered
3. An exterior angle and its adjacent interior angle form a linear pair (supplementary)
4. The sum of all exterior angles (one at each vertex) = 360°
5. An exterior angle is formed by extending one side of the triangle
1. Each triangle has 6 exterior angles (two at each vertex)
2. Only one exterior angle at each vertex is typically considered
3. An exterior angle and its adjacent interior angle form a linear pair (supplementary)
4. The sum of all exterior angles (one at each vertex) = 360°
5. An exterior angle is formed by extending one side of the triangle
Example 1: Find exterior angle
In △ABC, ∠A = 50° and ∠B = 60°. Find the exterior angle at C.
Using Exterior Angle Theorem:
$\text{Exterior angle at C} = \angle A + \angle B$
$\text{Exterior angle at C} = 50° + 60°$
$\text{Exterior angle at C} = 110°$
Alternative method using angle sum:
First find ∠C: $50° + 60° + \angle C = 180°$ → $\angle C = 70°$
Exterior angle at C = $180° - 70° = 110°$ ✓
Answer: Exterior angle = 110°
In △ABC, ∠A = 50° and ∠B = 60°. Find the exterior angle at C.
Using Exterior Angle Theorem:
$\text{Exterior angle at C} = \angle A + \angle B$
$\text{Exterior angle at C} = 50° + 60°$
$\text{Exterior angle at C} = 110°$
Alternative method using angle sum:
First find ∠C: $50° + 60° + \angle C = 180°$ → $\angle C = 70°$
Exterior angle at C = $180° - 70° = 110°$ ✓
Answer: Exterior angle = 110°
Example 2: Find interior angle using exterior angle
The exterior angle of a triangle is 130°. If one remote interior angle is 75°, find the other remote interior angle.
Using Exterior Angle Theorem:
$\text{Exterior angle} = \text{Remote angle 1} + \text{Remote angle 2}$
$130° = 75° + \text{Remote angle 2}$
$\text{Remote angle 2} = 130° - 75°$
$\text{Remote angle 2} = 55°$
Answer: Other remote interior angle = 55°
The exterior angle of a triangle is 130°. If one remote interior angle is 75°, find the other remote interior angle.
Using Exterior Angle Theorem:
$\text{Exterior angle} = \text{Remote angle 1} + \text{Remote angle 2}$
$130° = 75° + \text{Remote angle 2}$
$\text{Remote angle 2} = 130° - 75°$
$\text{Remote angle 2} = 55°$
Answer: Other remote interior angle = 55°
Example 3: Algebraic application
In △ABC, ∠A = x, ∠B = 2x, and the exterior angle at C = 120°. Find x and all interior angles.
Using Exterior Angle Theorem:
$\angle A + \angle B = \text{Exterior angle at C}$
$x + 2x = 120°$
$3x = 120°$
$x = 40°$
Find all angles:
$\angle A = x = 40°$
$\angle B = 2x = 80°$
$\angle C = 180° - (40° + 80°) = 60°$
Verify: Exterior angle at C = $180° - 60° = 120°$ ✓
Answer: x = 40°, ∠A = 40°, ∠B = 80°, ∠C = 60°
In △ABC, ∠A = x, ∠B = 2x, and the exterior angle at C = 120°. Find x and all interior angles.
Using Exterior Angle Theorem:
$\angle A + \angle B = \text{Exterior angle at C}$
$x + 2x = 120°$
$3x = 120°$
$x = 40°$
Find all angles:
$\angle A = x = 40°$
$\angle B = 2x = 80°$
$\angle C = 180° - (40° + 80°) = 60°$
Verify: Exterior angle at C = $180° - 60° = 120°$ ✓
Answer: x = 40°, ∠A = 40°, ∠B = 80°, ∠C = 60°
Proof of Exterior Angle Theorem:
Consider △ABC with side BC extended to D, forming exterior angle ∠ACD.
Step 1: ∠ACB + ∠ACD = 180° (linear pair)
Step 2: ∠A + ∠B + ∠ACB = 180° (Triangle Angle-Sum Theorem)
Step 3: From steps 1 and 2: ∠ACB + ∠ACD = ∠A + ∠B + ∠ACB
Step 4: Subtract ∠ACB from both sides: ∠ACD = ∠A + ∠B
Therefore, the exterior angle equals the sum of the two remote interior angles.
Consider △ABC with side BC extended to D, forming exterior angle ∠ACD.
Step 1: ∠ACB + ∠ACD = 180° (linear pair)
Step 2: ∠A + ∠B + ∠ACB = 180° (Triangle Angle-Sum Theorem)
Step 3: From steps 1 and 2: ∠ACB + ∠ACD = ∠A + ∠B + ∠ACB
Step 4: Subtract ∠ACB from both sides: ∠ACD = ∠A + ∠B
Therefore, the exterior angle equals the sum of the two remote interior angles.
4. Exterior Angle Inequality
Exterior Angle Inequality Theorem: Compares exterior angle to remote interior angles
Key Concept: The exterior angle is GREATER than either individual remote interior angle
Use: To establish relationships and prove inequalities in triangles
Key Concept: The exterior angle is GREATER than either individual remote interior angle
Use: To establish relationships and prove inequalities in triangles
Exterior Angle Inequality Theorem:
Statement: The measure of an exterior angle of a triangle is greater than the measure of either of the remote (non-adjacent) interior angles.
For triangle ABC with exterior angle ∠d at vertex C:
$$\angle d > \angle A \quad \text{AND} \quad \angle d > \angle B$$
In general form:
$$\text{Exterior Angle} > \text{Either Remote Interior Angle}$$
Note: The exterior angle is greater than EACH remote interior angle individually, but equals their SUM.
Statement: The measure of an exterior angle of a triangle is greater than the measure of either of the remote (non-adjacent) interior angles.
For triangle ABC with exterior angle ∠d at vertex C:
$$\angle d > \angle A \quad \text{AND} \quad \angle d > \angle B$$
In general form:
$$\text{Exterior Angle} > \text{Either Remote Interior Angle}$$
Note: The exterior angle is greater than EACH remote interior angle individually, but equals their SUM.
Understanding the Inequality:
Given exterior angle = $\angle A + \angle B$ (Exterior Angle Theorem):
Since angles are positive:
• $\angle A + \angle B > \angle A$ (because $\angle B > 0$)
• $\angle A + \angle B > \angle B$ (because $\angle A > 0$)
Therefore:
• Exterior angle > ∠A
• Exterior angle > ∠B
This is true for ALL six exterior angles of a triangle!
Given exterior angle = $\angle A + \angle B$ (Exterior Angle Theorem):
Since angles are positive:
• $\angle A + \angle B > \angle A$ (because $\angle B > 0$)
• $\angle A + \angle B > \angle B$ (because $\angle A > 0$)
Therefore:
• Exterior angle > ∠A
• Exterior angle > ∠B
This is true for ALL six exterior angles of a triangle!
Example 1: Apply exterior angle inequality
In △PQR, the exterior angle at R is 115°. What can you conclude about ∠P and ∠Q?
By Exterior Angle Inequality Theorem:
• Exterior angle at R > ∠P
• Exterior angle at R > ∠Q
Therefore:
• $115° > \angle P$ → $\angle P < 115°$
• $115° > \angle Q$ → $\angle Q < 115°$
Also by Exterior Angle Theorem:
$\angle P + \angle Q = 115°$
Conclusion: Each of ∠P and ∠Q is less than 115°, and together they sum to 115°
In △PQR, the exterior angle at R is 115°. What can you conclude about ∠P and ∠Q?
By Exterior Angle Inequality Theorem:
• Exterior angle at R > ∠P
• Exterior angle at R > ∠Q
Therefore:
• $115° > \angle P$ → $\angle P < 115°$
• $115° > \angle Q$ → $\angle Q < 115°$
Also by Exterior Angle Theorem:
$\angle P + \angle Q = 115°$
Conclusion: Each of ∠P and ∠Q is less than 115°, and together they sum to 115°
Example 2: Determine validity
Can a triangle have an exterior angle of 80° and a remote interior angle of 85°?
By Exterior Angle Inequality Theorem:
Exterior angle must be GREATER than any remote interior angle
Given: Exterior angle = 80°, Remote interior angle = 85°
But 80° < 85° (NOT greater than)
Answer: NO, this is impossible. It violates the Exterior Angle Inequality Theorem.
Can a triangle have an exterior angle of 80° and a remote interior angle of 85°?
By Exterior Angle Inequality Theorem:
Exterior angle must be GREATER than any remote interior angle
Given: Exterior angle = 80°, Remote interior angle = 85°
But 80° < 85° (NOT greater than)
Answer: NO, this is impossible. It violates the Exterior Angle Inequality Theorem.
Example 3: Find range of values
In △ABC, ∠A = 40° and ∠B = 70°. If x is the measure of an exterior angle at C, what is the range of x?
By Exterior Angle Theorem:
$x = \angle A + \angle B = 40° + 70° = 110°$
By Exterior Angle Inequality:
$x > \angle A$ → $x > 40°$
$x > \angle B$ → $x > 70°$
In this case, x has a specific value of 110°
Verify: 110° > 40° ✓ and 110° > 70° ✓
Answer: x = 110°, which satisfies x > 40° and x > 70°
In △ABC, ∠A = 40° and ∠B = 70°. If x is the measure of an exterior angle at C, what is the range of x?
By Exterior Angle Theorem:
$x = \angle A + \angle B = 40° + 70° = 110°$
By Exterior Angle Inequality:
$x > \angle A$ → $x > 40°$
$x > \angle B$ → $x > 70°$
In this case, x has a specific value of 110°
Verify: 110° > 40° ✓ and 110° > 70° ✓
Answer: x = 110°, which satisfies x > 40° and x > 70°
Example 4: Proof application
Given: In △XYZ, exterior angle at Z = (3x + 20)°, ∠X = (x + 10)°, ∠Y = (2x - 5)°
Prove: The exterior angle is greater than each remote interior angle.
First, find x using Exterior Angle Theorem:
$(3x + 20) = (x + 10) + (2x - 5)$
$3x + 20 = 3x + 5$
$20 = 5$ ← This is inconsistent!
Let's use correct setup:
Exterior angle = $3x + 20$
$\angle X = x + 10$
$\angle Y = 2x - 5$
Using theorem: $3x + 20 = (x + 10) + (2x - 5)$
$3x + 20 = 3x + 5$
This means the problem has inconsistent values. For a valid problem:
If exterior angle = 110°, ∠X = 40°, ∠Y = 70°
Then: 110° > 40° ✓ and 110° > 70° ✓
Given: In △XYZ, exterior angle at Z = (3x + 20)°, ∠X = (x + 10)°, ∠Y = (2x - 5)°
Prove: The exterior angle is greater than each remote interior angle.
First, find x using Exterior Angle Theorem:
$(3x + 20) = (x + 10) + (2x - 5)$
$3x + 20 = 3x + 5$
$20 = 5$ ← This is inconsistent!
Let's use correct setup:
Exterior angle = $3x + 20$
$\angle X = x + 10$
$\angle Y = 2x - 5$
Using theorem: $3x + 20 = (x + 10) + (2x - 5)$
$3x + 20 = 3x + 5$
This means the problem has inconsistent values. For a valid problem:
If exterior angle = 110°, ∠X = 40°, ∠Y = 70°
Then: 110° > 40° ✓ and 110° > 70° ✓
Triangle Classification Summary
| Classification Type | Triangle Type | Properties | Example |
|---|---|---|---|
| By Sides | Equilateral | All 3 sides equal, all angles 60° | 5, 5, 5 |
| Isosceles | Exactly 2 sides equal, 2 angles equal | 5, 5, 7 | |
| Scalene | All sides different, all angles different | 3, 4, 5 | |
| By Angles | Acute | All angles < 90° | 60°, 70°, 50° |
| Right | One angle = 90° | 90°, 60°, 30° | |
| Obtuse | One angle > 90° | 110°, 40°, 30° | |
| Equiangular | All angles equal (60° each) | 60°, 60°, 60° |
Triangle Theorems Quick Reference
| Theorem | Statement | Formula |
|---|---|---|
| Triangle Angle-Sum Theorem | Sum of interior angles equals 180° | $\angle A + \angle B + \angle C = 180°$ |
| Exterior Angle Theorem | Exterior angle = sum of remote interior angles | $\angle \text{ext} = \angle 1 + \angle 2$ |
| Exterior Angle Inequality | Exterior angle > either remote interior angle | $\angle \text{ext} > \angle 1$ AND $\angle \text{ext} > \angle 2$ |
| Linear Pair (Supplementary) | Exterior and adjacent interior angles | $\angle \text{ext} + \angle \text{adj} = 180°$ |
Special Triangle Properties
| Triangle Type | Side Property | Angle Property | Special Note |
|---|---|---|---|
| Equilateral | All sides equal | All angles = 60° | Also equiangular and acute |
| Isosceles | Two sides equal | Base angles equal | Has line of symmetry |
| Right | Follows Pythagorean Theorem | One angle = 90°, two acute | Acute angles are complementary |
| Right Isosceles | Two legs equal | 45°-45°-90° | Special right triangle |
Angle Relationships
| Relationship | Definition | Formula/Property |
|---|---|---|
| Interior Angles | Angles inside the triangle | Sum = 180° |
| Exterior Angle | Angle outside triangle, formed by extension | = Sum of two remote interior angles |
| Adjacent Interior Angle | Interior angle next to exterior angle | Forms linear pair with exterior angle |
| Remote Interior Angles | Two interior angles not adjacent to exterior | Each is less than exterior angle |
| Sum of All Exterior Angles | One at each vertex | Always = 360° |
Success Tips for Triangle Theorems:
✓ All interior angles of ANY triangle sum to 180°
✓ Equilateral triangles have all sides AND angles equal (60° each)
✓ Isosceles triangles have two equal sides and two equal base angles
✓ Right triangles have one 90° angle; the other two are complementary
✓ Exterior Angle = sum of two REMOTE interior angles
✓ Exterior Angle > EACH remote interior angle individually
✓ Exterior angle and adjacent interior angle are supplementary (180°)
✓ Sum of all exterior angles (one per vertex) = 360°
✓ To find missing angle: Use angle sum = 180°
✓ Always check your answer makes sense (angles between 0° and 180°)!
✓ All interior angles of ANY triangle sum to 180°
✓ Equilateral triangles have all sides AND angles equal (60° each)
✓ Isosceles triangles have two equal sides and two equal base angles
✓ Right triangles have one 90° angle; the other two are complementary
✓ Exterior Angle = sum of two REMOTE interior angles
✓ Exterior Angle > EACH remote interior angle individually
✓ Exterior angle and adjacent interior angle are supplementary (180°)
✓ Sum of all exterior angles (one per vertex) = 360°
✓ To find missing angle: Use angle sum = 180°
✓ Always check your answer makes sense (angles between 0° and 180°)!