Difference between revisions of "Menelaus' Theorem"

Revision as of 23:01, 10 March 2008

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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 $P, Q, R$ on the respective sides $BC, CA, AB$ (or their extensions) of a triangle $ABC$ to be collinear is that

$BP\cdot CQ\cdot AR = -PC\cdot QA\cdot RB$

where all segments in the formula are directed segments.

$[asy] defaultpen(fontsize(8)); pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R; draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R); label("A",A,(1,1));label("B",B,(-1,0));label("C",C,(1,0));label("P",P,(0,-1));label("Q",Q,(1,0));label("R",R,(-1,1)); [/asy]$

Proof

Draw a line parallel to $QP$ through $A$ to intersect $BC$ at $K$ :

$[asy] defaultpen(fontsize(8)); pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R, K=(5.5,0); draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); draw(A--K, dashed); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R^^K); label("A",A,(1,1));label("B",B,(-1,0));label("C",C,(1,0));label("P",P,(0,-1));label("Q",Q,(1,0));label("R",R,(-1,1)); label("K",K,(0,-1)); [/asy]$

$\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}$

$\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{PC}{PK}$

Multiplying the two equalities together to eliminate the $PK$ factor, we get:

$\frac{AR}{RB}\cdot\frac{QC}{QA}=\frac{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}=1$

@@ Line 4: / Line 4: @@
 It is named for Menelaus of Alexandria.
 == Statement ==
-A necessary and sufficient condition for points <math>D, E, F</math> on the respective side lines <math>BC, CA, AB</math> 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>BD\cdot CE\cdot AF = -DC\cdot EA\cdot FB</math></center>
+<center><math>BP\cdot CQ\cdot AR = -PC\cdot QA\cdot RB</math></center>
 where all segments in the formula are [[directed segment]]s.
-[[Image:Menelaus1.PNG|center]]
+<center><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></center>
+== Proof ==
+Draw a line parallel to <math>QP</math> through <math>A</math> to intersect <math>BC</math> at <math>K</math>:
+<center><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></center>
+<math>\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}</math>
+<math>\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{PC}{PK}</math>
+Multiplying the two equalities together to eliminate the <math>PK</math> factor, we get:
+<math>\frac{AR}{RB}\cdot\frac{QC}{QA}=\frac{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}=1</math>
 == See also ==
 * [[Ceva's Theorem]]

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Difference between revisions of "Menelaus' Theorem"

Revision as of 23:01, 10 March 2008

Statement

Proof

See also