Ceva's Theorem/Proof 2

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

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.

$\blacksquare$