# Ceva's theorem

(Redirected from Ceva's Theorem)

Ceva's Theorem (pronounced "Cheva") is a very popular theorem in elementary geometry. Given a triangle ABC, and points D, E, and F that lie on lines BC, CA, and AB respectively, the theorem states that lines AD, BE and CF are concurrent if and only if

[itex]\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.[itex]

It was first proved by Giovanni Ceva.

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Cevastheorem.jpg

## Proof

Suppose [itex]AD[itex], [itex]BE[itex] and [itex]CF[itex] intersect at a point [itex]X[itex]. Because [itex]\triangle BXD[itex] and [itex]\triangle CXD[itex] have the same height, we have

[itex]\frac{|\triangle BXD|}{|\triangle CXD|}=\frac{BD}{DC}.[itex]

Similarly,

From this it follows that

[itex]\frac{BD}{DC}=

=\frac{|\triangle ABX|}{|\triangle CAX|}.[itex]

Similarly,

[itex]\frac{CE}{EA}=\frac{|\triangle BCX|}{|\triangle ABX|}[itex], and
[itex]\frac{AF}{FB}=\frac{|\triangle CAX|}{|\triangle BCX|}[itex].

Multiplying these three equations gives

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

as required. Conversely, suppose that the points [itex]D[itex], [itex]E[itex] and [itex]F[itex] satisfy the above equality. Let [itex]AD[itex] and [itex]BE[itex] intersect at [itex]X[itex], and let [itex]CX[itex] intersect [itex]AB[itex] at [itex]F'[itex]. By the direction we have just proven,

[itex]\frac{AF'}{F'B} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.[itex]

Comparing with the above equality, we obtain

[itex]\frac{AF'}{F'B}=\frac{AF}{FB}.[itex]

Adding 1 to both sides and using [itex]AF'+F'B=AF+FB=AB[itex], we obtain

[itex]\frac{AB}{F'B}=\frac{AB}{FB}.[itex]

Thus [itex]F'B=FB[itex], so that [itex]F[itex] and [itex]F'[itex] coincide (recalling that the distances are directed). Therefore [itex]AD[itex], [itex]BE[itex] and [itex]CF[itex]=[itex]CF'[itex] intersect at [itex]X[itex], and both implications are proven.

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