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Ceva's Theorem

Revision as of 15:43, 24 December 2006 by Boy Soprano II (talk | contribs) (Missing dollar sign; didn't actually revise content)

Ceva's Theorem is an algebraic statement regarding the lengths of cevians in a triangle.


Statement

A necessary and sufficient condition for $AD, BE, CF,$ where $D, E,$ and $F$ are points of the respective side lines $BC, CA, AB$ of a triangle $ABC$, to be concurrent is that


$BD\cdot CE\cdot AF = DC \cdot EA \cdot FB$


where all segments in the formula are directed segments.

Ceva1.PNG

Proof

Letting the altitude from $A$ to $BC$ have length $h$ we have $[ABD]=\frac 12 BD\cdot h$ and $[ACD]=\frac 12 DC\cdot h$ where the brackets represent area. Thus $\frac{[ABD]}{[ACD]} = \frac{BD}{DC}$. In the same manner, we find that $\frac{[XBD]}{[XCD]} = \frac{BD}{DC}$. Thus

$\frac{BD}{DC} = \frac{[ABD]}{[ACD]} = \frac{[XBD]}{[XCD]} = \frac{[ABD]-[XBD]}{[ACD]-[XCD]} = \frac{[ABX]}{[ACX]}.$

Likewise, we find that

$\frac{CE}{EA}$ $=\frac{[BCX]}{[ABX]}$
$\frac{AF}{FB}$ $=\frac{[ACX]}{[BCX]}$

Thus

$\frac{BD}{DC}\cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = \frac{[ABX]}{[ACX]}\cdot \frac{[BCX]}{[ABX]}\cdot \frac{[ACX]}{[BCX]} = 1 \Rightarrow BD\cdot CE\cdot AF = DC \cdot EA \cdot FB.$

$\mathcal{QED}$

Alternate Formulation

The trig version of Ceva's Theorem states that cevians $AD,BE,CF$ are concurrent if and only if

$\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA.$

Proof

This proof is incomplete. If you can finish it, please do so. Thanks!

We will use Ceva's Theorem in the form that was already proven to be true.

First, we show that if $\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA$, holds true then $BD\cdot CE\cdot AF = DC \cdot EA \cdot FB$ which gives that the cevians are concurrent by Ceva's Theorem. The Law of Sines tells us that

$\frac{BD}{\sin BAD} = \frac{AB}{\sin ADB} \Leftrightarrow \sin BAD = \frac{BD}{AB\sin ADB}.$

Likewise, we get

$\sin ACF = \frac{AF}{AC\sin CFA}$
$\sin CBE = \frac{CE}{BC\sin BEC}$
$\sin CAD = \frac{CD}{AC\sin ADC}$
$\sin BCF = \frac{BF}{BC\sin BFC}$
$\sin ABE = \frac{AE}{AB\sin AEB}$

Thus

$\sin BAD \sin ACF \sin CBE = \sin CAD \sin BCF \sin ABE$
$\frac{BD}{AB\sin ADB} \cdot \frac{AF}{AC\sin CFA} \cdot \frac{CE}{BC\sin BEC} = \frac{CD}{AC\sin ADC} \cdot \frac{BF}{BC\sin BFC} \cdot \frac{AE}{AB\sin AEB}$

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. See the proof of the concurrency of the altitudes of a triangle at the orthocenter.
  3. See the proof of the concurrency of the perpendicual bisectors of a triangle at the circumcenter.

See also

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