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Difference between revisions of "Ceva's Theorem"

(restated theorem in conventional (and, in my opinion, superior) way; rewrote first proof; also wrote proof for trig version (I think Law of Sines is a dead end))
(→‎Trigonometric Form: typos)
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and similarly,
 
and similarly,
 
<center>
 
<center>
−
<math> \frac{[CBX]}{[XBA]} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} </math>. </center>
+
<math> \frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} </math>. </center>
 
It follows that
 
It follows that
 
<center>
 
<center>
−
<math> \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{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}  </math> <br> <br>  <math> \qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1 </math>.
+
<math> \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}  </math> <br> <br>  <math> \qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1 </math>.
 
</center>
 
</center>
 +
 +
Here, sign is irrelevant, as we may interpret the sines of [[directed angles]] mod <math> \displaystyle \pi </math> 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.  {{Halmos}}
 
The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem.  {{Halmos}}

Revision as of 07:55, 8 June 2007

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


Statement

Let $\displaystyle ABC$ be a triangle, and let $\displaystyle D, E, F$ be points on lines $\displaystyle BC, CA, AB$, respectively. Lines $\displaystyle AD, BE, CF$ concur iff if and only if


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


where lengths are directed.

Ceva1.PNG

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

Proof

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

First, suppose $\displaystyle AD, BE, CF$ meet at a point $\displaystyle X$. We note that triangles $\displaystyle ABD, ADC$ have the same altitude to line $\displaystyle BC$, but bases $\displaystyle BD$ and $\displaystyle DC$. It follows that $\frac {BD}{DC} = \frac{[ABD]}{[ADC]}$. The same is true for triangles $\displaystyle XBD, XDC$, 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 $\displaystyle D, E,F$ satisfy Ceva's criterion, and suppose $\displaystyle AD, BE$ intersect at $\displaystyle X$. Suppose the line $\displaystyle CX$ intersects line $\displaystyle AB$ at $\displaystyle F'$. We have proven that $\displaystyle F'$ must satisfy Ceva's criterion. This means that

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

so

$\displaystyle F' = F$,

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

Trigonometric Form

The trigonometric form of Ceva's Theorem (Trig Ceva) states that cevians $\displaystyle AD,BE,CF$ 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 $\displaystyle AD, BE, CF$ concur at a point $\displaystyle X$. 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 $\displaystyle \pi$ 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. ∎

Examples

  1. Suppose AB, AC, and BC have lengths 13, 14, and 15. If $\frac{AF}{FB} = \frac{2}{5}$ and $\frac{CE}{EA} = \frac{5}{8}$. Find BD and DC.

    If $BD = x$ and $DC = y$, then $10x = 40y$, and ${x + y = 15}$. From this, we find $x = 12$ and $y = 3$.
  2. The concurrence of the altitudes of a triangle at the orthocenter and the concurrence of the perpendicual bisectors of a triangle at the circumcenter can both be proven by Ceva's Theorem (the latter is a little harder). Furthermore, the existance of the centroid can be shown by Ceva, and the existance 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 classic Ceva's Theorem can be proven using ratios of areas.
  3. The existance of isotonic conjugates can be shown by classic Ceva, and the existance of isogonal conjugates can be shown by trig Ceva.

See also

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