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 :


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

Proof

 * Cevas-Theorem.png

Necessary Condition
Let $AL$, $BM$ and $CN$ are 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 cancel out, leaving us with:


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

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.

Also see

 * Menelaus's Theorem