|
|
| (39 intermediate revisions by 19 users not shown) |
| Line 1: |
Line 1: |
| − | '''Ceva's Theorem''' is an algebraic statement regarding the lengths of [[Cevian|cevians]] in a [[triangle]].
| + | #REDIRECT[[Ceva's theorem]] |
| − | | |
| − | | |
| − | == Statement ==
| |
| − | ''(awaiting image)''<br>
| |
| − | 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
| |
| − | <br><center><math>BD * CE * AF = +DC * EA * FB</math></center><br>
| |
| − | where all segments in the formula are directed segments.
| |
| − | | |
| − | == Proof ==
| |
| − | 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>
| |
| − | The triangles <math>\displaystyle{\triangle{ABX}}</math> and <math>\displaystyle{\triangle{A'CX}}</math> are similar, and so are <math>\displaystyle\triangle{ABY}</math> and <math>\triangle{CB'Y}</math>. Then the following equalities hold:
| |
| − | <center><math>\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}</math></center>
| |
| − | <br>
| |
| − | and thus
| |
| − | <center><math>\frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C} \qquad(1)</math></center>
| |
| − | <br>
| |
| − | Notice that if directed segments are being used, then <math>AB</math> and <math>BA</math> have opposite signs, and therefore when cancelled change the sign of the expression. That's why we changed <math>CA'</math> to <math>A'C</math>.
| |
| − | <br><br>
| |
| − | Now we turn to consider the following similarities: <math>\triangle{AZP}\sim\triangle{A'CP}</math> and <math>\triangle BZP\sim\triangle B'CP</math>. From them we get the equalities
| |
| − | <center><math>\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}</math></center>
| |
| − | <br>
| |
| − | which lead to
| |
| − | <center><math>\frac{AZ}{ZB}=\frac{A'C}{CB'}</math>.</center>
| |
| − | <br>
| |
| − | Multiplying the last expression with (1) gives
| |
| − | <center><math>\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1</math></center>
| |
| − | <br>
| |
| − | and we conclude the proof.
| |
| − | <br><br>
| |
| − | To prove the converse, suppose that <math>{X,Y,Z}</math> are points on <math>{BC}, {CA}, {AB}</math> respectively and satisfying
| |
| − | <center><math>\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.</math></center>
| |
| − | <br>
| |
| − | Let <math>Q</math> be the intersection point of <math>AX</math> with <math>BY</math>, and let <math>Z'</math> be the intersection of <math>CQ</math> with <math>AB</math>. Since then <math>AX,BY,CZ'</math> are concurrent, we have
| |
| − | <center><math>\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1</math></center>
| |
| − | <br>
| |
| − | and thus
| |
| − | <center><math>\frac{AZ'}{Z'B}=\frac{AZ}{ZB}</math></center>
| |
| − | <br>
| |
| − | which implies <math>Z=Z'</math>, and therefore <math>AX,BY,CZ</math> are concurrent.
| |
| − | | |
| − | <div align="right">''(proof courtesy planetmath.org, used under GNU License)''</div>
| |
| − | | |
| − | == Example ==
| |
| − | Suppose AB, AC, and BC have lengths 13, 14, and 15. If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>. Find BD and DC.<br>
| |
| − | <br>
| |
| − | If <math>BD = x</math> and <math>DC = y</math>, then <math>10x = 40y</math>, and <math>{x + y = 15}</math>. From this, we find <math>x = 12</math> and <math>y = 3</math>.
| |
| − | | |
| − | == See also ==
| |
| − | * [[Menelaus' Theorem]]
| |
| − | * [[Stewart's Theorem]]
| |
