Ceva's Theorem

A proof of Ceva's theorem using area ratios: three cevians AD, BE, CF are concurrent if and only if (BD/DC)·(CE/EA)·(AF/FB) = 1.

Keywords: Ceva's theorem, concurrency, cevian, area ratios

Prerequisites: area-proof · Difficulty: intermediate

Three cevians $A D$ , $B E$ , $C F$ of triangle $A B C$ are concurrent if and only if the product of the side ratios they cut equals $1$ :

\frac{B D}{D C} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B} = 1

Strategy: prove the forward direction with area ratios. Each ratio $B D / D C$ equals $[△ A B P] / [△ A C P]$ for the concurrency point $P$ ; multiplying all three cancels every area, leaving the product $1$ .

Step 1: What Is a Cevian?

A cevian of triangle $A B C$ is a segment from a vertex to a point on the opposite side. For example, pick any point $D$ on side $B C$ ; the segment $A D$ is a cevian from vertex $A$ .

Medians, altitudes, and angle bisectors are all familiar cevians.

Step 2: Three Concurrent Cevians

Now pick points $E$ on side $C A$ and $F$ on side $A B$ , and draw cevians $B E$ and $C F$ . When the three cevians $A D$ , $B E$ , $C F$ pass through a single common point $P$ , we say they are concurrent.

Ceva's theorem will tell us exactly when this happens.

Step 3: Ceva's Theorem

Ceva's Theorem (1678): three cevians $A D$ , $B E$ , $C F$ of triangle $A B C$ are concurrent if and only if the product of the three side ratios equals $1$ :

\frac{B D}{D C} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B} = 1

We'll prove the forward direction (concurrent ⇒ product $= 1$ ) using area ratios.

Step 4: Area Ratios

Key lemma: two triangles with a common apex and bases on the same line have areas in the ratio of their bases — they share the same altitude from the apex.

Apply the lemma to $△ A B D$ and $△ A C D$ : they share apex $A$ , with bases $B D$ and $D C$ on line $B C$ . So:

\frac{[ △ A B D ]}{[ △ A C D ]} = \frac{B D}{D C}

The same lemma with apex $P$ (in place of $A$ ) gives $[P B D] / [P C D] = B D / D C$ . Subtracting the two equal ratios:

\frac{[ △ A B P ]}{[ △ A C P ]} = \frac{[ A B D ] - [ P B D ]}{[ A C D ] - [ P C D ]} = \frac{B D}{D C}

This is the form we need for the telescoping product below.

Step 5: Chain All Three Ratios

Similarly:

$C E / E A = [△ B C P] / [△ B A P]$
$A F / F B = [△ C A P] / [△ C B P]$

When we multiply all three ratios, every area appears once in the numerator and once in the denominator.

Step 6: Product $= 1$

Multiplying the three ratios:

\frac{B D}{D C} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B} = \frac{[ A B P ]}{[ A C P ]} \cdot \frac{[ B C P ]}{[ B A P ]} \cdot \frac{[ C A P ]}{[ C B P ]} = 1

Every area cancels! The product is exactly $1$ .

Step 7: Verification with Known Cevians

Medians: $D$ , $E$ , $F$ are midpoints, so each ratio is $1$ . Product $= 1 \cdot 1 \cdot 1 = 1$ . ✓

Altitudes: The ratios involve trigonometric expressions, but their product is still $1$ . ✓

Angle bisectors: By the angle bisector theorem, each ratio involves the adjacent sides. The product simplifies to $1$ . ✓

Ceva's Theorem: Cevians $A D$ , $B E$ , $C F$ are concurrent if and only if:

\frac{B D}{D C} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B} = 1

This unifies the concurrency of medians ( $= 1$ trivially), altitudes, and angle bisectors under a single criterion. ∎