Difference between revisions of "Menelaus' Theorem"
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A necessary and sufficient condition for points <math>P, Q, R</math> on the respective sides <math>BC, CA, AB</math> (or their extensions) of a triangle <math>ABC</math> to be collinear is that | A necessary and sufficient condition for points <math>P, Q, R</math> on the respective sides <math>BC, CA, AB</math> (or their extensions) of a triangle <math>ABC</math> to be collinear is that | ||
− | <center><math>BP\cdot CQ\cdot AR = | + | <center><math>BP\cdot CQ\cdot AR = PC\cdot QA\cdot RB</math></center> |
where all segments in the formula are [[directed segment]]s. | where all segments in the formula are [[directed segment]]s. |
Revision as of 13:29, 26 July 2010
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Menelaus' Theorem deals with the collinearity of points on each of the three sides (extended when necessary) of a triangle. It is named for Menelaus of Alexandria.
Statement
A necessary and sufficient condition for points on the respective sides
(or their extensions) of a triangle
to be collinear is that
![$BP\cdot CQ\cdot AR = PC\cdot QA\cdot RB$](http://latex.artofproblemsolving.com/a/9/9/a99776e14df29ff7c409db31dc3b74690fc0a610.png)
where all segments in the formula are directed segments.
![[asy] defaultpen(fontsize(8)); pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R; draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R); label("A",A,(1,1));label("B",B,(-1,0));label("C",C,(1,0));label("P",P,(0,-1));label("Q",Q,(1,0));label("R",R,(-1,1)); [/asy]](http://latex.artofproblemsolving.com/8/6/a/86a254a29864a11196f5bbd900856e311e41be6b.png)
Proof
Draw a line parallel to through
to intersect
at
:
![[asy] defaultpen(fontsize(8)); pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R, K=(5.5,0); draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); draw(A--K, dashed); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R^^K); label("A",A,(1,1));label("B",B,(-1,0));label("C",C,(1,0));label("P",P,(0,-1));label("Q",Q,(1,0));label("R",R,(-1,1)); label("K",K,(0,-1)); [/asy]](http://latex.artofproblemsolving.com/f/b/d/fbd7bf9ba4bcb21c083f9f6aef5d813523b36c8f.png)
Multiplying the two equalities together to eliminate the factor, we get: