Difference between revisions of "Menelaus' Theorem"
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− | '''Menelaus' Theorem''' deals with the [[collinearity]] of points on each of the three sides (extended when necessary) of a [[triangle]]. | + | '''Menelaus's 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. | It is named for Menelaus of Alexandria. | ||
== Statement == | == Statement == |
Revision as of 22:46, 28 May 2013
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Menelaus's 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/9/7/4/974c07d36b8fe13f0b58f5d2ac02809f9c78a1ec.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: