MenelausPlease see the following picture:
Triangle ABC is intersected by a line in points P,Q,R:
The theorem of Menelaus is:
Draw perpendicular lines : A-->A' B-->B' and C-->C'
This yields similar triangles AA'Q and CC'Q also CC'R and BB'R also AA'P and BB'P
If the product of line ratio's equals 1, the points P,Q,R are on a straight line.
The proof is not presented here.
Starting at point A, walk over the triangle perimeter in counterclockwise direction.
So the path is AB, BC and CA.
In case of AB, write A. as numerator and B. as denominator of a fraction.
Repeat for BC and CA:
For A./B. substitute P for (.)
For B./C. substitute R for (.)
For C./A. substitute Q for (.)
If three lines intersecting the angles of a triangle have a common point (S) then:
This is the theorem of Ceva.
We use the theorem of Menelaus.
Please look at ΔARC intersected by BQ and also observe ΔRBC intersected by AP.
If the product equals 1 the lines (crossing A,B,C) have a common point.
The sine rule of Ceva
This rule is:
if intersecting lines (crossing A,B,C) have a common point (S).
Repeated application of the trigonometric sine rule in triangles ABS, BCS and ASC yields:
If the above sine ratio's products are true, the lines through angles A,B,C intersect in one point.