Properties of Triangles - Tenth Grade Geometry
1. Midsegments of Triangles
Midsegment: A line segment connecting the midpoints of two sides of a triangle
Also Called: Midline
Number of Midsegments: Every triangle has exactly 3 midsegments
Key Property: Parallel to the third side and half its length
Also Called: Midline
Number of Midsegments: Every triangle has exactly 3 midsegments
Key Property: Parallel to the third side and half its length
Triangle Midsegment Theorem:
If a segment connects the midpoints of two sides of a triangle, then:
1. The midsegment is PARALLEL to the third side
$$DE \parallel BC$$
2. The midsegment is HALF the length of the third side
$$DE = \frac{1}{2}BC$$
Where D is the midpoint of AB and E is the midpoint of AC in △ABC
If a segment connects the midpoints of two sides of a triangle, then:
1. The midsegment is PARALLEL to the third side
$$DE \parallel BC$$
2. The midsegment is HALF the length of the third side
$$DE = \frac{1}{2}BC$$
Where D is the midpoint of AB and E is the midpoint of AC in △ABC
Properties of Midsegments:
1. Creates Similar Triangle:
The midsegment creates a smaller triangle similar to the original triangle
2. Area Relationship:
The smaller triangle has $\frac{1}{4}$ the area of the original triangle
3. Perimeter Relationship:
The smaller triangle has $\frac{1}{2}$ the perimeter of the original triangle
4. Four Congruent Triangles:
The three midsegments divide the original triangle into 4 congruent triangles
1. Creates Similar Triangle:
The midsegment creates a smaller triangle similar to the original triangle
2. Area Relationship:
The smaller triangle has $\frac{1}{4}$ the area of the original triangle
3. Perimeter Relationship:
The smaller triangle has $\frac{1}{2}$ the perimeter of the original triangle
4. Four Congruent Triangles:
The three midsegments divide the original triangle into 4 congruent triangles
Example 1: Find midsegment length
In △ABC, D and E are midpoints of AB and AC respectively
If BC = 18, find DE
By Midsegment Theorem:
$$DE = \frac{1}{2}BC = \frac{1}{2}(18) = 9$$
Answer: DE = 9 units
In △ABC, D and E are midpoints of AB and AC respectively
If BC = 18, find DE
By Midsegment Theorem:
$$DE = \frac{1}{2}BC = \frac{1}{2}(18) = 9$$
Answer: DE = 9 units
Example 2: Find third side
If midsegment MN = 12, find the length of the parallel side
Using Midsegment Theorem:
$$MN = \frac{1}{2}(\text{Third side})$$
$$12 = \frac{1}{2}(\text{Third side})$$
$$\text{Third side} = 24$$
Answer: Third side = 24 units
If midsegment MN = 12, find the length of the parallel side
Using Midsegment Theorem:
$$MN = \frac{1}{2}(\text{Third side})$$
$$12 = \frac{1}{2}(\text{Third side})$$
$$\text{Third side} = 24$$
Answer: Third side = 24 units
2. Triangles and Bisectors
Angle Bisector
Angle Bisector: A ray that divides an angle into two equal parts
Property: Divides the angle into two congruent angles
In Triangle: Goes from vertex to opposite side
Property: Divides the angle into two congruent angles
In Triangle: Goes from vertex to opposite side
Angle Bisector Theorem:
An angle bisector of a triangle divides the opposite side into segments proportional to the other two sides.
If AD bisects ∠A in △ABC, then:
$$\frac{BD}{DC} = \frac{AB}{AC}$$
Where D is on side BC
An angle bisector of a triangle divides the opposite side into segments proportional to the other two sides.
If AD bisects ∠A in △ABC, then:
$$\frac{BD}{DC} = \frac{AB}{AC}$$
Where D is on side BC
Perpendicular Bisector
Perpendicular Bisector: A line perpendicular to a segment at its midpoint
Property: Every point on the perpendicular bisector is equidistant from the endpoints
Forms: Two right angles (90°) at the midpoint
Property: Every point on the perpendicular bisector is equidistant from the endpoints
Forms: Two right angles (90°) at the midpoint
Perpendicular Bisector Theorem:
If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
$$PA = PB$$
Where P is any point on the perpendicular bisector of AB
Converse: If a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector
If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
$$PA = PB$$
Where P is any point on the perpendicular bisector of AB
Converse: If a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector
3. Identify Medians, Altitudes, Angle Bisectors, and Perpendicular Bisectors
Median
Median: A line segment from a vertex to the midpoint of the opposite side
Every Triangle Has: Exactly 3 medians
Key Property: All three medians intersect at one point (the centroid)
Function: Divides triangle into two triangles of equal area
Every Triangle Has: Exactly 3 medians
Key Property: All three medians intersect at one point (the centroid)
Function: Divides triangle into two triangles of equal area
Properties of Medians:
• Each median divides the triangle into two smaller triangles of equal area
• The three medians divide the triangle into six smaller triangles of equal area
• The length of a median can be calculated using:
$$m_a = \frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$$
Where $m_a$ is the median to side $a$, and $b, c$ are the other two sides
• Each median divides the triangle into two smaller triangles of equal area
• The three medians divide the triangle into six smaller triangles of equal area
• The length of a median can be calculated using:
$$m_a = \frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$$
Where $m_a$ is the median to side $a$, and $b, c$ are the other two sides
Altitude
Altitude: A perpendicular line segment from a vertex to the opposite side (or its extension)
Every Triangle Has: Exactly 3 altitudes
Key Property: Forms a right angle with the base
Also Called: Height
Every Triangle Has: Exactly 3 altitudes
Key Property: Forms a right angle with the base
Also Called: Height
Location of Altitudes:
Acute Triangle:
• All three altitudes lie inside the triangle
Right Triangle:
• Two altitudes are the legs of the triangle
• Third altitude is inside the triangle
Obtuse Triangle:
• Two altitudes lie outside the triangle
• One altitude is inside the triangle
Acute Triangle:
• All three altitudes lie inside the triangle
Right Triangle:
• Two altitudes are the legs of the triangle
• Third altitude is inside the triangle
Obtuse Triangle:
• Two altitudes lie outside the triangle
• One altitude is inside the triangle
Comparison Table
| Type | Definition | Starts From | Ends At | Special Property |
|---|---|---|---|---|
| Median | Connects vertex to midpoint of opposite side | Vertex | Midpoint of opposite side | Divides triangle into equal areas |
| Altitude | Perpendicular from vertex to opposite side | Vertex | Opposite side (⊥) | Forms 90° angle |
| Angle Bisector | Divides angle into two equal parts | Vertex | Opposite side | Creates two equal angles |
| Perpendicular Bisector | Perpendicular to side at its midpoint | Anywhere | Midpoint of side (⊥) | Equidistant from endpoints |
4. Angle-Side Relationships in Triangles
Key Concept: In any triangle, there is a direct relationship between the sizes of angles and lengths of opposite sides
Main Idea: Larger angle = longer opposite side; Smaller angle = shorter opposite side
Main Idea: Larger angle = longer opposite side; Smaller angle = shorter opposite side
Angle-Side Inequality Theorem:
1. If one side is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side.
$$\text{If } a > b, \text{ then } \angle A > \angle B$$
2. If one angle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle.
$$\text{If } \angle A > \angle B, \text{ then } a > b$$
1. If one side is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side.
$$\text{If } a > b, \text{ then } \angle A > \angle B$$
2. If one angle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle.
$$\text{If } \angle A > \angle B, \text{ then } a > b$$
Key Relationships:
In any triangle:
• The longest side is opposite the largest angle
• The shortest side is opposite the smallest angle
• The medium side is opposite the medium angle
Order Preservation:
The order of side lengths matches the order of opposite angles
In any triangle:
• The longest side is opposite the largest angle
• The shortest side is opposite the smallest angle
• The medium side is opposite the medium angle
Order Preservation:
The order of side lengths matches the order of opposite angles
Example 1: Order angles by size
In △ABC, AB = 7, BC = 10, AC = 5
Order the angles from smallest to largest
Order sides: AC < AB < BC (5 < 7 < 10)
Order opposite angles: ∠B < ∠C < ∠A
Answer: ∠B is smallest, ∠A is largest
In △ABC, AB = 7, BC = 10, AC = 5
Order the angles from smallest to largest
Order sides: AC < AB < BC (5 < 7 < 10)
Order opposite angles: ∠B < ∠C < ∠A
Answer: ∠B is smallest, ∠A is largest
Example 2: Order sides by length
In △PQR, ∠P = 40°, ∠Q = 70°, ∠R = 70°
Order the sides from shortest to longest
Order angles: ∠P < ∠Q = ∠R (40° < 70° = 70°)
Order opposite sides: QR < PR = PQ
Answer: QR is shortest, PR and PQ are equal (isosceles triangle)
In △PQR, ∠P = 40°, ∠Q = 70°, ∠R = 70°
Order the sides from shortest to longest
Order angles: ∠P < ∠Q = ∠R (40° < 70° = 70°)
Order opposite sides: QR < PR = PQ
Answer: QR is shortest, PR and PQ are equal (isosceles triangle)
5. Triangle Inequality Theorem
Triangle Inequality Theorem: A fundamental rule about which sets of three lengths can form a triangle
Key Idea: The sum of any two sides must be greater than the third side
Purpose: Determines if three given lengths can form a triangle
Key Idea: The sum of any two sides must be greater than the third side
Purpose: Determines if three given lengths can form a triangle
Triangle Inequality Theorem:
For any triangle with sides $a$, $b$, and $c$:
$$a + b > c$$
$$b + c > a$$
$$a + c > b$$
In words: The sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
All three inequalities must be true for a triangle to exist!
For any triangle with sides $a$, $b$, and $c$:
$$a + b > c$$
$$b + c > a$$
$$a + c > b$$
In words: The sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
All three inequalities must be true for a triangle to exist!
Practical Application:
To check if three lengths can form a triangle:
1. Add the two smallest sides
2. Compare to the largest side
3. If sum > largest side, triangle can exist
4. If sum ≤ largest side, triangle CANNOT exist
Shortcut: Just check if (smallest + middle) > largest
To check if three lengths can form a triangle:
1. Add the two smallest sides
2. Compare to the largest side
3. If sum > largest side, triangle can exist
4. If sum ≤ largest side, triangle CANNOT exist
Shortcut: Just check if (smallest + middle) > largest
Example 1: Can these form a triangle?
Sides: 5, 7, 10
Check: $5 + 7 = 12 > 10$ ✓
(We only need to check the sum of the two smaller sides)
Answer: YES, these can form a triangle
Sides: 5, 7, 10
Check: $5 + 7 = 12 > 10$ ✓
(We only need to check the sum of the two smaller sides)
Answer: YES, these can form a triangle
Example 2: Can these form a triangle?
Sides: 3, 4, 8
Check: $3 + 4 = 7 < 8$ ✗
Answer: NO, these cannot form a triangle (sum of two sides not greater than third)
Sides: 3, 4, 8
Check: $3 + 4 = 7 < 8$ ✗
Answer: NO, these cannot form a triangle (sum of two sides not greater than third)
Example 3: Find possible range for third side
Two sides of a triangle are 5 and 12. Find the range of possible values for the third side x.
Using Triangle Inequality:
$5 + 12 > x$ → $x < 17$
$5 + x > 12$ → $x > 7$
$12 + x > 5$ → $x > -7$ (always true for positive x)
Answer: 7 < x < 17 (third side must be between 7 and 17)
Two sides of a triangle are 5 and 12. Find the range of possible values for the third side x.
Using Triangle Inequality:
$5 + 12 > x$ → $x < 17$
$5 + x > 12$ → $x > 7$
$12 + x > 5$ → $x > -7$ (always true for positive x)
Answer: 7 < x < 17 (third side must be between 7 and 17)
6. Construct the Circumcenter or Incenter of a Triangle
Circumcenter
Circumcenter: The point where the three perpendicular bisectors of the sides intersect
Symbol: Usually denoted as O
Special Property: Equidistant from all three vertices
Center of: Circumscribed circle (circumcircle)
Symbol: Usually denoted as O
Special Property: Equidistant from all three vertices
Center of: Circumscribed circle (circumcircle)
Circumcenter Properties:
Definition: Intersection of perpendicular bisectors
Distance Property:
$$OA = OB = OC = R$$
Where O is the circumcenter and R is the circumradius
Location:
• Acute triangle: Inside the triangle
• Right triangle: On the hypotenuse (midpoint)
• Obtuse triangle: Outside the triangle
Definition: Intersection of perpendicular bisectors
Distance Property:
$$OA = OB = OC = R$$
Where O is the circumcenter and R is the circumradius
Location:
• Acute triangle: Inside the triangle
• Right triangle: On the hypotenuse (midpoint)
• Obtuse triangle: Outside the triangle
Steps to Construct Circumcenter:
Step 1: Draw perpendicular bisector of side AB
Step 2: Draw perpendicular bisector of side BC
Step 3: Find the intersection point O
Step 4: O is the circumcenter (equidistant from A, B, and C)
Step 5: Draw circle with center O and radius OA (circumcircle)
Step 1: Draw perpendicular bisector of side AB
Step 2: Draw perpendicular bisector of side BC
Step 3: Find the intersection point O
Step 4: O is the circumcenter (equidistant from A, B, and C)
Step 5: Draw circle with center O and radius OA (circumcircle)
Incenter
Incenter: The point where the three angle bisectors intersect
Symbol: Usually denoted as I
Special Property: Equidistant from all three sides
Center of: Inscribed circle (incircle)
Location: Always inside the triangle
Symbol: Usually denoted as I
Special Property: Equidistant from all three sides
Center of: Inscribed circle (incircle)
Location: Always inside the triangle
Incenter Properties:
Definition: Intersection of angle bisectors
Distance Property:
The incenter is equidistant from all three sides (perpendicular distance)
$$d(I, AB) = d(I, BC) = d(I, AC) = r$$
Where I is the incenter and r is the inradius
Location: Always inside the triangle (for all types)
Definition: Intersection of angle bisectors
Distance Property:
The incenter is equidistant from all three sides (perpendicular distance)
$$d(I, AB) = d(I, BC) = d(I, AC) = r$$
Where I is the incenter and r is the inradius
Location: Always inside the triangle (for all types)
Steps to Construct Incenter:
Step 1: Draw angle bisector of ∠A
Step 2: Draw angle bisector of ∠B
Step 3: Find the intersection point I
Step 4: I is the incenter (equidistant from all three sides)
Step 5: Draw perpendicular from I to any side to find radius r
Step 6: Draw circle with center I and radius r (incircle)
Step 1: Draw angle bisector of ∠A
Step 2: Draw angle bisector of ∠B
Step 3: Find the intersection point I
Step 4: I is the incenter (equidistant from all three sides)
Step 5: Draw perpendicular from I to any side to find radius r
Step 6: Draw circle with center I and radius r (incircle)
7. Construct the Centroid or Orthocenter of a Triangle
Centroid
Centroid: The point where the three medians intersect
Symbol: Usually denoted as G
Also Called: Center of gravity or center of mass
Special Property: Balance point of the triangle
Location: Always inside the triangle
Symbol: Usually denoted as G
Also Called: Center of gravity or center of mass
Special Property: Balance point of the triangle
Location: Always inside the triangle
Centroid Properties:
Definition: Intersection of medians
2:1 Ratio Property:
The centroid divides each median in the ratio 2:1 from vertex to midpoint
$$\frac{AG}{GD} = \frac{BG}{GE} = \frac{CG}{GF} = \frac{2}{1}$$
Where D, E, F are midpoints and G is the centroid
Coordinate Formula:
If vertices are A$(x_1, y_1)$, B$(x_2, y_2)$, C$(x_3, y_3)$, then:
$$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$$
Definition: Intersection of medians
2:1 Ratio Property:
The centroid divides each median in the ratio 2:1 from vertex to midpoint
$$\frac{AG}{GD} = \frac{BG}{GE} = \frac{CG}{GF} = \frac{2}{1}$$
Where D, E, F are midpoints and G is the centroid
Coordinate Formula:
If vertices are A$(x_1, y_1)$, B$(x_2, y_2)$, C$(x_3, y_3)$, then:
$$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$$
Steps to Construct Centroid:
Step 1: Find midpoint D of side BC
Step 2: Draw median from A to D
Step 3: Find midpoint E of side AC
Step 4: Draw median from B to E
Step 5: Find intersection point G (centroid)
Step 6: Verify: AG is twice the length of GD
Step 1: Find midpoint D of side BC
Step 2: Draw median from A to D
Step 3: Find midpoint E of side AC
Step 4: Draw median from B to E
Step 5: Find intersection point G (centroid)
Step 6: Verify: AG is twice the length of GD
Orthocenter
Orthocenter: The point where the three altitudes intersect
Symbol: Usually denoted as H
Special Property: Intersection of perpendicular lines from vertices
Location: Depends on triangle type
Symbol: Usually denoted as H
Special Property: Intersection of perpendicular lines from vertices
Location: Depends on triangle type
Orthocenter Properties:
Definition: Intersection of altitudes
Location:
• Acute triangle: Inside the triangle
• Right triangle: At the right angle vertex
• Obtuse triangle: Outside the triangle
Euler Line:
The orthocenter (H), centroid (G), and circumcenter (O) are always collinear
$$HG:GO = 2:1$$
Definition: Intersection of altitudes
Location:
• Acute triangle: Inside the triangle
• Right triangle: At the right angle vertex
• Obtuse triangle: Outside the triangle
Euler Line:
The orthocenter (H), centroid (G), and circumcenter (O) are always collinear
$$HG:GO = 2:1$$
Steps to Construct Orthocenter:
Step 1: From vertex A, draw altitude perpendicular to BC
Step 2: From vertex B, draw altitude perpendicular to AC
Step 3: Find intersection point H (orthocenter)
Step 4: Verify by drawing third altitude from C to AB
Note: May need to extend sides for obtuse triangles
Step 1: From vertex A, draw altitude perpendicular to BC
Step 2: From vertex B, draw altitude perpendicular to AC
Step 3: Find intersection point H (orthocenter)
Step 4: Verify by drawing third altitude from C to AB
Note: May need to extend sides for obtuse triangles
8. Find the Centroid of a Triangle
Centroid Coordinate Formula:
For triangle with vertices A$(x_1, y_1)$, B$(x_2, y_2)$, C$(x_3, y_3)$:
$$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$$
In words: Average of the x-coordinates, average of the y-coordinates
For triangle with vertices A$(x_1, y_1)$, B$(x_2, y_2)$, C$(x_3, y_3)$:
$$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$$
In words: Average of the x-coordinates, average of the y-coordinates
Example 1: Find centroid coordinates
Triangle with vertices A(0, 0), B(6, 0), C(3, 9)
Find the centroid G
$$G_x = \frac{0 + 6 + 3}{3} = \frac{9}{3} = 3$$
$$G_y = \frac{0 + 0 + 9}{3} = \frac{9}{3} = 3$$
Answer: Centroid G(3, 3)
Triangle with vertices A(0, 0), B(6, 0), C(3, 9)
Find the centroid G
$$G_x = \frac{0 + 6 + 3}{3} = \frac{9}{3} = 3$$
$$G_y = \frac{0 + 0 + 9}{3} = \frac{9}{3} = 3$$
Answer: Centroid G(3, 3)
Example 2: Using 2:1 ratio
In △ABC, median AD = 12. Find AG and GD where G is centroid.
Centroid divides median in 2:1 ratio:
$$AG:GD = 2:1$$
$$AG = \frac{2}{3} \times 12 = 8$$
$$GD = \frac{1}{3} \times 12 = 4$$
Answer: AG = 8, GD = 4
In △ABC, median AD = 12. Find AG and GD where G is centroid.
Centroid divides median in 2:1 ratio:
$$AG:GD = 2:1$$
$$AG = \frac{2}{3} \times 12 = 8$$
$$GD = \frac{1}{3} \times 12 = 4$$
Answer: AG = 8, GD = 4
Triangle Centers Summary
| Center | Intersection Of | Equidistant From | Location | Center Of |
|---|---|---|---|---|
| Centroid (G) | 3 Medians | - | Always inside | Balance point |
| Circumcenter (O) | 3 Perpendicular bisectors | 3 Vertices | Inside/on/outside | Circumcircle |
| Incenter (I) | 3 Angle bisectors | 3 Sides | Always inside | Incircle |
| Orthocenter (H) | 3 Altitudes | - | Inside/on/outside | - |
Triangle Center Locations
| Triangle Type | Centroid (G) | Circumcenter (O) | Incenter (I) | Orthocenter (H) |
|---|---|---|---|---|
| Acute | Inside | Inside | Inside | Inside |
| Right | Inside | On hypotenuse | Inside | At right angle |
| Obtuse | Inside | Outside | Inside | Outside |
| Equilateral | All four centers coincide at the same point | |||
Key Formulas Quick Reference
| Concept | Formula/Property | Use |
|---|---|---|
| Midsegment | $DE = \frac{1}{2}BC$ and $DE \parallel BC$ | Find midsegment or third side |
| Angle Bisector Theorem | $\frac{BD}{DC} = \frac{AB}{AC}$ | Find proportional segments |
| Triangle Inequality | $a + b > c$, $b + c > a$, $a + c > b$ | Check if triangle exists |
| Centroid Coordinates | $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$ | Find centroid location |
| Centroid Ratio | $AG:GD = 2:1$ | Find distances on median |
| Angle-Side Relationship | Larger angle ↔ Longer opposite side | Order angles/sides |
Segment Comparison
| Property | Median | Altitude | Angle Bisector | Perpendicular Bisector |
|---|---|---|---|---|
| Starts from | Vertex | Vertex | Vertex | Side |
| Goes to | Midpoint of opposite side | Opposite side (⊥) | Opposite side | Midpoint of side (⊥) |
| Angle formed | Any | 90° | Two equal angles | 90° |
| Intersection point | Centroid (G) | Orthocenter (H) | Incenter (I) | Circumcenter (O) |
| Number in triangle | 3 | 3 | 3 | 3 |
Important Theorems Summary
| Theorem | Statement | Application |
|---|---|---|
| Midsegment Theorem | Midsegment is parallel and half the third side | Find unknown lengths |
| Triangle Inequality | Sum of two sides > third side | Determine if triangle exists |
| Angle-Side Inequality | Larger angle opposite longer side | Order sides or angles |
| Angle Bisector Theorem | Divides opposite side proportionally | Find segment lengths |
| Perpendicular Bisector | Equidistant from endpoints | Find circumcenter |
| Centroid 2:1 Ratio | Divides median 2:1 from vertex | Find distances on medians |
Success Tips for Triangle Properties:
✓ Midsegment: parallel to third side and half its length
✓ Triangle Inequality: sum of any two sides > third side
✓ Angle-Side: larger angle ↔ longer opposite side
✓ Four centers: Centroid (medians), Circumcenter (⊥ bisectors), Incenter (angle bisectors), Orthocenter (altitudes)
✓ Centroid divides each median 2:1 from vertex to midpoint
✓ Centroid coordinates: average of x's, average of y's
✓ Circumcenter equidistant from vertices; Incenter equidistant from sides
✓ Median connects vertex to midpoint; Altitude is perpendicular to side
✓ Angle bisector divides angle equally; Perpendicular bisector divides side equally
✓ Euler line: H, G, and O are always collinear with ratio HG:GO = 2:1
✓ Midsegment: parallel to third side and half its length
✓ Triangle Inequality: sum of any two sides > third side
✓ Angle-Side: larger angle ↔ longer opposite side
✓ Four centers: Centroid (medians), Circumcenter (⊥ bisectors), Incenter (angle bisectors), Orthocenter (altitudes)
✓ Centroid divides each median 2:1 from vertex to midpoint
✓ Centroid coordinates: average of x's, average of y's
✓ Circumcenter equidistant from vertices; Incenter equidistant from sides
✓ Median connects vertex to midpoint; Altitude is perpendicular to side
✓ Angle bisector divides angle equally; Perpendicular bisector divides side equally
✓ Euler line: H, G, and O are always collinear with ratio HG:GO = 2:1