Ceva's Theorem

A proof of Ceva's theorem using area ratios: three cevians AD, BE, CF are concurrent if and only if $\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB} = 1$.

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

Prerequisites: area-proof · Difficulty: intermediate

Ceva's Theorem (1678): Three cevians $AD$, $BE$, $CF$ of triangle $ABC$ are concurrent if and only if:

\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1
    

Proof Strategy (Area Ratios)

If the three cevians meet at point $P$, then for each ratio:

\frac{BD}{DC} = \frac{[\triangle ABP]}{[\triangle ACP]}
    

because triangles $ABP$ and $ACP$ share vertex $A$ (hence the same height from $A$ to line $BC$), so their areas are in the ratio of their bases $BD$ and $DC$.

Multiplying all three such ratios and using the cancellation of areas gives the product equal to $1$.

The Proof

Step 1: What Is a Cevian?

A cevian of triangle $ABC$ is a segment from a vertex to a point on the opposite side. For example, pick any point $D$ on side $BC$; the segment $AD$ 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 $CA$ and $F$ on side $AB$, and draw cevians $BE$ and $CF$. When the three cevians $AD$, $BE$, $CF$ 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 $AD$, $BE$, $CF$ of triangle $ABC$ are concurrent if and only if the product of the three side ratios equals $1$:

\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 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 $\triangle ABD$ and $\triangle ACD$: they share apex $A$, with bases $BD$ and $DC$ on line $BC$. So:

\frac{[\triangle ABD]}{[\triangle ACD]} = \frac{BD}{DC}
    

The same lemma with apex $P$ (in place of $A$) gives $[PBD]/[PCD] = BD/DC$. Subtracting the two equal ratios:

\frac{[\triangle ABP]}{[\triangle ACP]}
    = \frac{[ABD] - [PBD]}{[ACD] - [PCD]} = \frac{BD}{DC}
    

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

Step 5: Chain All Three Ratios

Similarly:

• $CE/EA = [\triangle BCP] / [\triangle BAP]$
• $AF/FB = [\triangle CAP] / [\triangle CBP]$

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{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB}
    = \frac{[ABP]}{[ACP]} \cdot \frac{[BCP]}{[BAP]} \cdot \frac{[CAP]}{[CBP]} = 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 $1$. ✓

Conclusion

Ceva's Theorem: Cevians $AD$, $BE$, $CF$ are concurrent if and only if:

\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1
    

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