Ceva's theorem

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For other uses, see Ceva (Disambiguation).

Ceva's theorem, case 1: the three lines are concurrent at a point O inside ABC

Ceva's theorem, case 2: the three lines are concurrent at a point O outside ABC

Ceva's theorem 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

$\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.$

There is also an equivalent trigonometric form of Ceva's Theorem, that is, AD,BE,CF concur if and only if
$\frac{\sin\angle BAD}{\sin\angle CAD}\times\frac{\sin\angle ACF}{\sin\angle BCF}\times\frac{\sin\angle CBE}{\sin\angle ABE}=1$ .

It was first proven by Giovanni Ceva in his 1678 work De lineis rectis.

A Cevian is a segment which joins a vertex of a triangle with a point on the opposite side.

[edit] Proof of the theorem

Suppose $A D$ , $B E$ and $C F$ intersect at a point $O$ . Because $\triangle BOD$ and $\triangle COD$ have the same height, we have

$\frac{|\triangle BOD|}{|\triangle COD|}=\frac{BD}{DC}.$

Similarly,

$\frac{|\triangle BAD|}{|\triangle CAD|}=\frac{BD}{DC}.$

From this it follows that

$\frac{BD}{DC}= \frac{|\triangle BAD|-|\triangle BOD|}{|\triangle CAD|-|\triangle COD|} =\frac{|\triangle ABO|}{|\triangle CAO|}.$

Similarly,

$\frac{CE}{EA}=\frac{|\triangle BCO|}{|\triangle ABO|},$

and

$\frac{AF}{FB}=\frac{|\triangle CAO|}{|\triangle BCO|}.$

Multiplying these three equations gives

$\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1,$

as required. Conversely, suppose that the points $D$ , $E$ and $F$ satisfy the above equality. Let $A D$ and $B E$ intersect at $O$ , and let $C O$ intersect $A B$ at $F'$ . By the direction we have just proven,

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

Comparing with the above equality, we obtain

$\frac{AF'}{F'B}=\frac{AF}{FB}.$

Adding 1 to both sides and using $A F'' + F'' B = A F + F B = A B$ , we obtain

$\frac{AB}{F'B}=\frac{AB}{FB}.$

Thus $F'' B = F B$ , so that $F$ and $F''$ coincide (recalling that the distances are directed). Therefore $A D$ , $B E$ and $C F$ = $C F''$ intersect at $O$ , and both implications are proven.