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 $L$.

This is a contradiction.

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

The result follows.

$\blacksquare$

Also see

• Menelaus's Theorem

Source of Name

This entry was named for Giovanni Benedetto Ceva.

Historical Note

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

Sources

• 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Ceva's theorem
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Ceva's theorem
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Ceva's theorem
• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Ceva's Theorem