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Difference between revisions of "Ceva's Theorem"

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===Intermediate===
 
===Intermediate===
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In <math>\Delta ABC, AD, BE, CF</math> are concurrent lines. <math>P, Q, R</math> are points on <math>EF, FD, DE</math> such that <math>DP, EQ, FR</math> are concurrent. Prove that (using ''plane geometry'') <math>AP, BQ, CR</math> are concurrent.<url>viewtopic.php?f=151&t=543574 [source + answer]<url>
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===Olympiad===
 
===Olympiad===
  

Revision as of 10:02, 14 July 2013

Ceva's Theorem is a criterion for the concurrence of cevians in a triangle.


Statement

Ceva1.PNG

Let $ABC$ be a triangle, and let $D, E, F$ be points on lines $BC, CA, AB$, respectively. Lines $AD, BE, CF$ are concurrent iff (if and only if)


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


where lengths are directed. This also works for the reciprocal or each of the ratios, as the reciprocal of $1$ is $1$.


(Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)


The proof using Routh's Theorem is extremely trivial, so we will not include it.

Proof

We will use the notation $[ABC]$ to denote the area of a triangle with vertices $A,B,C$.

First, suppose $AD, BE, CF$ meet at a point $X$. We note that triangles $ABD, ADC$ have the same altitude to line $BC$, but bases $BD$ and $DC$. It follows that $\frac {BD}{DC} = \frac{[ABD]}{[ADC]}$. The same is true for triangles $XBD, XDC$, so

$\frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]}$.

Similarly, $\frac{CE}{EA} = \frac{[BCX]}{[BXA]}$ and $\frac{AF}{FB} = \frac{[CAX]}{[CXB]}$, so

$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABX]}{[AXC]} \cdot \frac{[BCX]}{[BXA]} \cdot \frac{[CAX]}{[CXB]} = 1$.

Now, suppose $D, E,F$ satisfy Ceva's criterion, and suppose $AD, BE$ intersect at $X$. Suppose the line $CX$ intersects line $AB$ at $F'$. We have proven that $F'$ must satisfy Ceva's criterion. This means that

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

so

$F' = F$,

and line $CF$ concurrs with $AD$ and $BE$. ∎

Proof by Barycentric coordinates

Since $D\in BC$, we can write its coordinates as $(0,d,1-d)$. The equation of line $AD$ is then $z=\frac{1-d}{d}y$.

Similarly, since $E=(1-e,0,e)$, and $F=(f,1-f,0)$, we can see that the equations of $BE$ and $CF$ respectively are $x=\frac{1-e}{e}z$ and $y=\frac{1-f}{f}x$

Multiplying the three together yields the solution to the equation:

$xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y$

Dividing by $xyz$ yields:


$1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}$, which is equivalent to Ceva's theorem

QED

Trigonometric Form

The trigonometric form of Ceva's Theorem (Trig Ceva) states that cevians $AD,BE,CF$ concur if and only if

$\frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1.$

Proof

First, suppose $AD, BE, CF$ concur at a point $X$. We note that

$\frac{[BAX]}{[XAC]} = \frac{ \frac{1}{2}AB \cdot AX \cdot \sin BAX}{ \frac{1}{2}AX \cdot AC \cdot \sin XAC} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC}$,

and similarly,

$\frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}$.

It follows that

$\frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} \cdot \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}$

$\qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1$.

Here, sign is irrelevant, as we may interpret the sines of directed angles mod $\pi$ 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 $AB, AC$, and $BC$ have lengths $13, 14$, and $15$, respectively. If $\frac{AF}{FB} = \frac{2}{5}$ and $\frac{CE}{EA} = \frac{5}{8}$, find $BD$ and $DC$. (Source)

Intermediate

In $\Delta ABC, AD, BE, CF$ are concurrent lines. $P, Q, R$ are points on $EF, FD, DE$ such that $DP, EQ, FR$ are concurrent. Prove that (using plane geometry) $AP, BQ, CR$ are concurrent.<url>viewtopic.php?f=151&t=543574 [source + answer]<url>

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 isotomic conjugates can be shown by classic Ceva, and the existence of isogonal conjugates can be shown by trig Ceva.

See also

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