|
|
| (43 intermediate revisions by 18 users not shown) |
| Line 1: |
Line 1: |
| − | {{stub}}
| + | #REDIRECT[[Menelaus' theorem]] |
| − | | |
| − | '''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 <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 = -PC\cdot QA\cdot RB</math></center>
| |
| − | | |
| − | where all segments in the formula are [[directed segment]]s.
| |
| − | | |
| − | <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]]
| |
| − | * [[Stewart's Theorem]]
| |
| − | | |
| − | [[Category:Theorems]]
| |
| − | | |
| − | [[Category:Geometry]]
| |
