Editing Ceva's theorem (section)

===Using triangle areas===

First, the sign of the [[left-hand side]] is positive since either all three of the ratios are positive, the case where {{mvar|O}} is inside the triangle (upper diagram), or one is positive and the other two are negative, the case {{mvar|O}} is outside the triangle (lower diagram shows one case).

To check the magnitude, note that the area of a triangle of a given height is proportional to its base. So 
: <math>\frac{|\triangle BOD|}{|\triangle COD|}=\frac{\overline{BD}}{\overline{DC}}=\frac{|\triangle BAD|}{|\triangle CAD|}.</math>
Therefore,
:<math>\frac{\overline{BD}}{\overline{DC}}=
\frac{|\triangle BAD|-|\triangle BOD|}{|\triangle CAD|-|\triangle COD|}
=\frac{|\triangle ABO|}{|\triangle CAO|}.</math>
(Replace the minus with a plus if {{mvar|A}} and {{mvar|O}} are on opposite sides of {{mvar|BC}}.)
Similarly,
: <math>\frac{\overline{CE}}{\overline{EA}}=\frac{|\triangle BCO|}{|\triangle ABO|},</math>
and
: <math>\frac{\overline{AF}}{\overline{FB}}=\frac{|\triangle CAO|}{|\triangle BCO|}.</math>
Multiplying these three equations gives
: <math>\left|\frac{\overline{AF}}{\overline{FB}}  \cdot \frac{\overline{BD}}{\overline{DC}} \cdot \frac{\overline{CE}}{\overline{EA}} \right|= 1,</math>
as required.

The theorem can also be proven easily using [[Menelaus's theorem]].<ref>Follows {{cite book |title=Inductive Plane Geometry|url=https://archive.org/details/inductiveplanege00hopkrich|first=George Irving|last=Hopkins|publisher=D.C. Heath & Co.|year=1902|chapter=Art. 986}}</ref> From the transversal {{mvar|BOE}} of triangle {{math|△''ACF''}},
: <math>\frac{\overline{AB}}{\overline{BF}} \cdot \frac{\overline{FO}}{\overline{OC}} \cdot \frac{\overline{CE}}{\overline{EA}} = -1</math>
and from the transversal {{mvar|AOD}} of triangle {{math|△''BCF''}},
: <math>\frac{\overline{BA}}{\overline{AF}} \cdot \frac{\overline{FO}}{\overline{OC}} \cdot \frac{\overline{CD}}{\overline{DB}} = -1.</math>
The theorem follows by dividing these two equations.

The converse follows as a corollary.<ref name=r1/> Let {{mvar|D, E, F}} be given on the lines {{mvar|BC, AC, AB}} so that the equation holds. Let {{mvar|AD, BE}} meet at {{mvar|O}} and let {{mvar|F'}} be the point where {{mvar|CO}} crosses {{mvar|AB}}. Then by the theorem, the equation also holds for {{mvar|D, E, F'}}. Comparing the two, 
: <math>\frac{\overline{AF}}{\overline{FB}} = \frac{\overline{AF'}}{\overline{F'B}}</math>
But at most one point can cut a segment in a given ratio so {{mvar|1=F = F’}}.