Concurrency Theorems

The Secret Paths Inside Triangles

Discover the magical theorems that reveal hidden truths about points and lines

The Triangle Detective Story

Imagine you’re a detective. Your mission? Find where three secret paths meet inside a triangle. These paths could be anything—roads through a park, strings stretched across a frame, or even invisible laser beams in a spy movie.

Our analogy: Think of a triangle as a pizza with three edges. Sometimes you want to slice it in special ways, and certain slices meet at magical points. Today, we’ll learn the rules that tell us WHEN and WHERE these slices meet!

Ceva’s Theorem: When Three Slices Meet

The Pizza Parlor Problem

Picture this: You have a triangular pizza with corners A, B, and C. You want to cut three slices from each corner toward the opposite edge. Will your three cuts all meet at ONE point inside?

Ceva’s Theorem gives you the answer!

What It Says (Simple Version)

If you draw lines from each corner of a triangle to a point on the opposite side:

• From corner A to point D on side BC
• From corner B to point E on side CA
• From corner C to point F on side AB

These three lines meet at ONE point if and only if:

(AF/FB) × (BD/DC) × (CE/EA) = 1

The Magic Rule Explained

Think of it like a balance game:

• Each fraction shows how a point divides a side
• When you multiply all three fractions, you get exactly 1
• It’s like a perfect recipe—all ingredients in exact proportion!

graph TD
    A["Corner A"] --> D["Point D on BC"]
    B["Corner B"] --> E["Point E on CA"]
    C["Corner C"] --> F["Point F on AB"]
    D --> P["They all meet at P!"]
    E --> P
    F --> P

Real Example

Triangle ABC where:

• F divides AB so AF = 2 and FB = 3
• D divides BC so BD = 3 and DC = 2
• E divides CA so CE = 1 and EA = 1

Check: (2/3) × (3/2) × (1/1) = 1 ✓

Result: The three lines AD, BE, and CF all pass through the same point!

Menelaus’ Theorem: The Straight Line Cutter

A Different Puzzle

Now imagine someone takes a long straight stick and lays it across your triangle. This stick crosses all three sides (or their extensions). Where does it cross each side?

Menelaus’ Theorem tells you the relationship!

What It Says (Simple Version)

If a straight line cuts:

• Side BC (or its extension) at point D
• Side CA (or its extension) at point E
• Side AB (or its extension) at point F

Then:

(AF/FB) × (BD/DC) × (CE/EA) = -1

Wait, Why Negative?

The “-1” is like a warning sign! It tells you that the line crosses the triangle in a special way—at least one of the points is OUTSIDE the triangle’s edge (on the extension).

Think of it like this:

• Ceva = +1 (all paths meet inside, cozy gathering)
• Menelaus = -1 (the line slashes through, some points outside)

graph TD
    L["Straight Line"] --> D["Crosses BC at D"]
    L --> E["Crosses CA at E"]
    L --> F["Crosses AB at F"]
    D --> CHECK["Product = -1"]
    E --> CHECK
    F --> CHECK

Real Example

A line crosses triangle ABC:

• At F on side AB where AF = 4, FB = 2
• At D on extended BC where BD = 3, DC = -1 (outside!)
• At E on side CA where CE = 1, EA = 6

Check: (4/2) × (3/-1) × (1/6) = 2 × (-3) × (1/6) = -1 ✓

The negative DC tells us D is on the extension of BC, not between B and C.

Spotting the Difference

Applications of Ceva-Menelaus

Why Do We Care?

These theorems are like super-tools in your geometry toolkit!

Application 1: Finding Special Points

The Centroid (where medians meet):

• Medians connect each corner to the midpoint of the opposite side
• Each midpoint divides its side in ratio 1:1
• Check: (1/1) × (1/1) × (1/1) = 1 ✓
• Ceva confirms all medians meet at ONE point!

The Incenter (where angle bisectors meet):

• Angle bisectors divide opposite sides in the ratio of adjacent sides
• Ceva’s theorem confirms they meet at one point!

Application 2: Proving Collinearity

Want to prove three points lie on a straight line? Use Menelaus!

Example: In triangle ABC, show that certain points D, E, F are collinear by checking if the product equals -1.

Application 3: Competition Problems

These theorems appear in math olympiads constantly!

Typical problem structure:

1. Given: A triangle with points on sides
2. Find: Whether lines are concurrent OR points are collinear
3. Method: Calculate the Ceva or Menelaus product

graph TD
    P["Problem Given"] --> Q{What to prove?}
    Q --> |Lines meet at point| C["Use Ceva = 1"]
    Q --> |Points on same line| M["Use Menelaus = -1"]
    C --> S["Calculate ratios"]
    M --> S
    S --> V["Verify product"]

Application 4: Trigonometric Versions

For angle-based problems, there are trig versions:

Ceva (Trig):

sin(∠BAD)/sin(∠DAC) × sin(∠CBE)/sin(∠EBA) × sin(∠ACF)/sin(∠FCB) = 1

Menelaus (Trig): Similar pattern with sine ratios = -1

Stewart’s Theorem: The Distance Finder

A New Challenge

Forget about where lines meet. Now ask: How LONG is a line from a corner to the opposite side?

Stewart’s Theorem answers this!

The Setup

Draw a line from corner A to point D on side BC:

• Let BC = a (the whole base)
• Let BD = m and DC = n (so m + n = a)
• Let AD = d (the line we want to measure)
• Let AB = c and AC = b (the other two sides)

Stewart’s Formula

b²m + c²n = a(d² + mn)

Or rearranged:

d² = (b²m + c²n)/a - mn

The Memory Trick

“A man and his dad put a bomb in the sink.”

b²m + c²n = a(d² + mn)

• man = m, a, n
• dad = d, a, d
• bomb = b, m, b → b²m
• sink = c, n, c → c²n

Real Example

Triangle with AB = 5, AC = 7, BC = 8 Point D is on BC with BD = 3, DC = 5 Find AD.

Using Stewart’s:

• b = AC = 7, c = AB = 5, a = BC = 8
• m = BD = 3, n = DC = 5

b²m + c²n = a(d² + mn) 49(3) + 25(5) = 8(d² + 15) 147 + 125 = 8d² + 120 272 = 8d² + 120 152 = 8d² d² = 19 d = √19 ≈ 4.36

Answer: AD ≈ 4.36 units

Why Stewart’s Matters

1. Find median lengths (when D is the midpoint)
2. Find angle bisector lengths (when D divides BC by the angle bisector ratio)
3. Find altitude lengths (combine with other formulas)

graph TD
    S[Stewart's Theorem] --> M["Find Median Length"]
    S --> B["Find Angle Bisector Length"]
    S --> A["Find Altitude Length"]
    S --> G["General Cevian Length"]

Special Cases of Stewart’s

The Median Formula

When D is the midpoint of BC (so m = n = a/2):

d² = (2b² + 2c² - a²)/4

Or: Median to side a = (1/2)√(2b² + 2c² - a²)

The Angle Bisector Formula

When AD bisects angle A (so BD/DC = AB/AC = c/b):

d² = bc[(b+c)² - a²]/(b+c)²

Putting It All Together

The Geometry Detective’s Toolkit

Quick Decision Flow

graph TD
    Q["Your Question"] --> T1{About concurrent lines?}
    T1 --> |Yes| CEVA["Use Ceva = 1"]
    T1 --> |No| T2{About collinear points?}
    T2 --> |Yes| MEN["Use Menelaus = -1"]
    T2 --> |No| T3{About segment length?}
    T3 --> |Yes| STEW[Use Stewart's Theorem]
    T3 --> |No| OTHER["Use other tools"]

Practice Your Detective Skills

Example 1: Ceva in Action

In triangle ABC, lines AD, BE, CF meet at point P. If AF/FB = 2/3 and BD/DC = 3/4, find CE/EA.

Solution: (AF/FB) × (BD/DC) × (CE/EA) = 1 (2/3) × (3/4) × (CE/EA) = 1 (1/2) × (CE/EA) = 1 CE/EA = 2

Example 2: Menelaus in Action

Line L cuts triangle ABC at D (on BC), E (on CA), F (on AB). If AF/FB = 3/2 and BD/DC = 2/3, find CE/EA.

Solution: (AF/FB) × (BD/DC) × (CE/EA) = -1 (3/2) × (2/3) × (CE/EA) = -1 1 × (CE/EA) = -1 CE/EA = -1

The negative means E is on the extension of CA!

Example 3: Stewart in Action

Find the length of the median from A to BC in a triangle with AB = 6, AC = 8, BC = 10.

Solution: Using median formula: d² = (2b² + 2c² - a²)/4

• a = BC = 10, b = AC = 8, c = AB = 6

d² = (2×64 + 2×36 - 100)/4 d² = (128 + 72 - 100)/4 d² = 100/4 = 25 d = 5

The median has length 5!

Your Confidence Checklist

After reading this guide, you should feel confident saying:

• [ ] I know Ceva’s Theorem and when to use it (product = 1)
• [ ] I know Menelaus’ Theorem and when to use it (product = -1)
• [ ] I understand the difference between concurrent and collinear
• [ ] I can use Stewart’s Theorem to find segment lengths
• [ ] I know the special formulas for medians and angle bisectors

The Big Picture

These four tools—Ceva, Menelaus, their applications, and Stewart—are like having X-ray vision for triangles. They let you see invisible relationships:

1. Ceva sees where paths converge
2. Menelaus sees straight-line connections
3. Stewart measures the unmeasurable

You’re now equipped with the secret knowledge of triangle detectives. Go forth and solve!