# Ceva's Theorem

## Theorem

Let $\triangle ABC$ be a triangle.

Let $L$, $M$ and $N$ be points on the sides $BC$, $AC$ and $AB$ respectively.

Then the lines $AL$, $BM$ and $CN$ are concurrent if and only if:

$\dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB} = 1$

## Proof 1

### Necessary Condition

Let $AL$, $BM$ and $CN$ be concurrent.

Let the point of concurrency be $P$.

Consider the triangles $\triangle ALB$ and $\triangle ALC$.

They have the same altitude from the common base $BC$.

Hence:

$\dfrac {\map \Area {ALB} } {\map \Area {ALC} } = \dfrac {BL} {LC}$

Similarly, consider the triangles $\triangle PLB$ and $\triangle PLC$.

They also have the same altitude from the common base $BC$.

Hence:

$\dfrac {\map \Area {PLB} } {\map \Area {PLC} } = \dfrac {BL} {LC}$

Next we consider the triangles $\triangle APB$ and $\triangle APC$.

We have:

$\map \Area {APB} = \map \Area {ALB} - \map \Area {PLB}$

and:

$\map \Area {APC} = \map \Area {ALC} - \map \Area {PLC}$

and so:

$\dfrac {\map \Area {APB} } {\map \Area {APC} } = \dfrac {BL} {LC}$

In the same way, we derive:

$\dfrac {\map \Area {BPC} } {\map \Area {APB} } = \dfrac {CM} {MA}$

and:

$\dfrac {\map \Area {APC} } {\map \Area {BPC} } = \dfrac {AN} {NB}$

Thus we have:

$\dfrac {\map \Area {APB} } {\map \Area {APC} } \times \dfrac {\map \Area {BPC} } {\map \Area {APB} } \times \dfrac {\map \Area {APC} } {\map \Area {BPC} } = \dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB}$

The areas on the left hand side cancel out, leaving us with:

$\dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB} = 1$

$\Box$

### Sufficient Condition

Let:

$\dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB} = 1$

Let $P$ be the intersection of $AM$ and $CN$ as in the diagram above.

We need to show that $A$, $P$ and $L$ are collinear.

From the first part, we have that:

$\dfrac {\map \Area {BPC} } {\map \Area {APB} } = \dfrac {CM} {MA}$

and:

$\dfrac {\map \Area {APC} } {\map \Area {BPC} } = \dfrac {AN} {NB}$

Multiplying them:

$\dfrac {\map \Area {BPC} } {\map \Area {APB} } \times \dfrac {\map \Area {APC} } {\map \Area {BPC} } = \dfrac {CM} {MA} \times \dfrac {AN} {NB}$

Simplifying:

$\dfrac {\map \Area {APC} } {\map \Area {APB} } = \dfrac {CM} {MA} \times \dfrac {AN} {NB}$

It is given that:

$\dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB} = 1$

and so:

$\dfrac {CM} {MA} \times \dfrac {AN} {NB} = \dfrac {LC} {BL} = \dfrac {\map \Area {APC} } {\map \Area {APB} }$

Extend $BP$ to meet $AC$ at point $Z$, say.

By the same construction that we have used throughout, we have:

$\dfrac {\map \Area {APC} } {\map \Area {APB} } = \dfrac {ZC} {BZ}$

But then we have just shown that:

$\dfrac {LC} {BL} = \dfrac {\map \Area {APC} } {\map \Area {APB} }$

So $Z$ coincides with $L$ and the result follows.

$\blacksquare$

## Proof 2

### Necessary Condition

We have by hypothesis:

$AL$, $BM$ and $CN$ are concurrent in $\triangle ABC$ at point $P$.

Following the sides anticlockwise in $\triangle LAB$:

 $\text {(1)}: \quad$ $\ds \dfrac {LP} {PA} \cdot \dfrac {AN} {NB} \cdot \dfrac {BC} {CL}$ $=$ $\ds -1$ Menelaus's Theorem

Following the sides clockwise in $\triangle LAC$:

 $\text {(2)}: \quad$ $\ds \dfrac {LP} {PA} \cdot \dfrac {AM} {MC} \cdot \dfrac {CB} {BL}$ $=$ $\ds -1$ Menelaus's Theorem

Equate $(1)$ and $(2)$ and cancel $\dfrac {LP} {PA}$.

 $\ds \dfrac {AN} {NB} \cdot \dfrac {BC} {CL}$ $=$ $\ds \dfrac {AM} {MC} \cdot \dfrac {CB} {BL}$

By definition of directed line segments:

• $AM = -MA$
• $MC = -CM$
• $CL = -LC$
• $CB = -BC$

Hence:

 $\ds \dfrac {AN} {NB} \cdot \dfrac {BC} {-LC}$ $=$ $\ds \dfrac {-MA} {-CM} \cdot \dfrac {-BC} {BL}$ substituting $\ds \dfrac {AN} {NB} \cdot \dfrac {BC} {LC}$ $=$ $\ds \dfrac {MA} {CM} \cdot \dfrac {BC} {BL}$ cancel $-1$ twice $\ds \dfrac {CM} {MA} \cdot \dfrac {AN} {NB} \cdot \dfrac {BL} {LC}$ $=$ $\ds 1$ cancel $BC$ and rearrange

The result follows.

$\Box$

### Sufficient Condition

Given:

 $\ds \dfrac {CM} {MA} \cdot \dfrac {AN} {NB} \cdot \dfrac {BL} {LC}$ $=$ $\ds 1$ by hypothesis

Without loss of generality, let $BM$ and $CN$ be concurrent at $P$.

Suppose $AL$ does not go through point $P$.

Then let $AP$ produced to $BC$ to give $AL'$, where $L'$ is not the same point as $L$.

By the Necessary Condition:

 $\ds \dfrac {CM} {MA} \cdot \dfrac {AN} {NB} \cdot \dfrac {BL'} {L'C}$ $=$ $\ds 1$

Equating the two results:

 $\ds \dfrac {CM} {MA} \cdot \dfrac {AN} {NB} \cdot \dfrac {BL'} {L'C}$ $=$ $\ds \dfrac {CM} {MA} \cdot \dfrac {AN} {NB} \cdot \dfrac {BL} {LC}$ $\ds \leadsto \ \$ $\ds \dfrac {BL'} {L'C}$ $=$ $\ds \dfrac {BL} {LC}$ cancelling $\ds \dfrac {BL'} {L'C} + \dfrac {L'C} {L'C}$ $=$ $\ds \dfrac {BL} {LC} + \dfrac {LC} {LC}$ add $1$ to both sides $\ds \dfrac {BC} {L'C}$ $=$ $\ds \dfrac {BC} {LC}$ addition $\ds LC$ $=$ $\ds L'C$ cancel $BC$ and rearrange

Hence $L'$ is the same point as $L$.

$AL$, $BM$ and $CN$ are concurrent at $P$.

The result follows.

$\blacksquare$

## Source of Name

This entry was named for Giovanni Benedetto Ceva.

## Historical Note

Ceva's Theorem was discovered by Giovanni Benedetto Ceva in $1678$.