Difference between revisions of "Ceva's Theorem"
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− | Let <math>{X,Y,Z}</math> be points on <math>{BC}, {CA}, {AB}</math> respectively such that <math>AX,BY,CZ</math> are concurrent, and let <math>{P}</math> be the point where <math>AX</math>, <math>BY</math> and <math>CZ</math> meet. Draw a parallel to <math>AB</math> through the point <math>{C}</math>. Extend <math>AX</math> until it intersects the parallel at a point <math>{A'}</math>. Construct <math>\displaystyle{B'}</math> in a similar way extending <math>BY</math>. | + | Let <math>{X,Y,Z}</math> be points on <math>{BC}, {CA}, {AB}</math> respectively such that <math>AX,BY,CZ</math> are concurrent, and let <math>{P}</math> be the point where <math>AX</math>, <math>BY</math> and <math>CZ</math> meet. Draw a parallel to <math>AB</math> through the point <math>{C}</math>. Extend <math>AX</math> until it intersects the parallel at a point <math>\displaystyle{A'}</math>. Construct <math>\displaystyle{B'}</math> in a similar way extending <math>BY</math>. |
<center>''(ceva1.png)''</center> | <center>''(ceva1.png)''</center> | ||
− | The triangles <math>\triangle{ABX}<math> and <math>\triangle{A'CX}</math> are similar, and so are <math>\triangle{ABY}</math> and <math>\triangle{CB'Y}</math>. Then the following equalities hold: | + | The triangles <math>{\triangle{ABX}}</math> and <math>{\triangle{A'CX}}</math> are similar, and so are <math>\triangle{ABY}</math> and <math>\triangle{CB'Y}</math>. Then the following equalities hold: |
<math> | <math> | ||
Revision as of 15:38, 20 June 2006
Ceva's Theorem is an algebraic statement regarding the lengths of cevians in a triangle.
Contents
[hide]Statement
(awaiting image)
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
where all segments in the formula are directed segments.
Proof
Let be points on respectively such that are concurrent, and let be the point where , and meet. Draw a parallel to through the point . Extend until it intersects the parallel at a point . Construct in a similar way extending .
The triangles and are similar, and so are and . Then the following equalities hold:
$
and thus
$
Notice that if directed segments are being used then and have opposite signs, and therefore when cancelled change the sign of the expression. That's why we changed to .
Now we turn to consider the following similarities: and . From them we get the equalities
$
which lead to
$
Multiplying the last expression with (1) gives
$
and we conclude the proof.
To prove the converse, suppose that are points on respectively and satisfying
$
Let be the intersection point of with , and let be the intersection of with . Since then are concurrent, we have
$
and thus
$
which implies , and therefore are concurrent.
Example
Suppose AB, AC, and BC have lengths 13, 14, and 15. If and . Find BD and DC.
If and , then , and . From this, we find and .