Ceva's Theorem
Ceva's Theorem is a criterion for the concurrence of cevians in a triangle.
Contents
Statement
Let be a triangle, and let
be points on lines
, respectively. Lines
concur iff (if and only if)
,
where lengths are directed. This also works for the reciprocal or each of the ratios, as the reciprocal of is
.
(Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)
Proof
We will use the notation to denote the area of a triangle with vertices
.
First, suppose meet at a point
. We note that triangles
have the same altitude to line
, but bases
and
. It follows that
. The same is true for triangles
, so
![$\frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]}$](http://latex.artofproblemsolving.com/f/9/6/f96163bdaaebc0a8b7e5610b6ede75cd711d46c3.png)
Similarly, and
,
so
.
Now, suppose satisfy Ceva's criterion, and suppose
intersect at
. Suppose the line
intersects line
at
. We have proven that
must satisfy Ceva's criterion. This means that
![$\frac{AF'}{F'B} = \frac{AF}{FB}$](http://latex.artofproblemsolving.com/f/0/b/f0bd81e69ce6f73e83d1bfc12a702443e1de00ec.png)
so
![$F' = F$](http://latex.artofproblemsolving.com/f/1/3/f13301e6873d821edc62b5e33b80c0db4bff3c88.png)
and line concurrs with
and
. ∎
Trigonometric Form
The trigonometric form of Ceva's Theorem (Trig Ceva) states that cevians concur if and only if
Proof
First, suppose concur at a point
. We note that
![$\frac{[BAX]}{[XAC]} = \frac{ \frac{1}{2}AB \cdot AX \cdot \sin BAX}{ \frac{1}{2}AX \cdot AC \cdot \sin XAC} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC}$](http://latex.artofproblemsolving.com/8/8/1/881e383f76e1b1f48c4ce7c471a5a69b142de4c4.png)
and similarly,
![$\frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}$](http://latex.artofproblemsolving.com/c/8/c/c8c21fa17a564a7bc80e24391c269f03c8e97eeb.png)
It follows that
.
Here, sign is irrelevant, as we may interpret the sines of directed angles mod to be either positive or negative.
The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem. ∎
Problems
Introductory
- Suppose
, and
have lengths
, and
, respectively. If
and
, find
and
. (Source)
Intermediate
Olympiad
Other Notes
- The concurrence of the altitudes of a triangle at the orthocenter and the concurrence of the perpendicular bisectors of a triangle at the circumcenter can both be proven by Ceva's Theorem (the latter is a little harder). Furthermore, the existence of the centroid can be shown by Ceva, and the existence of the incenter can be shown using trig Ceva. However, there are more elegant methods for proving each of these results, and in any case, any result obtained by Ceva's Theorem can be obtained using ratios of areas.
- The existence of isotonic conjugates can be shown by classic Ceva, and the existence of isogonal conjugates can be shown by trig Ceva.