Menelaus, an ancient Greek mathematician, first proved Menelaus theorem. It is pointed out that if a straight line intersects with three sides AB, BC, CA of △ABC or its extension lines at points F, D and E, then AF/FB×BD/DC×CE/EA= 1.
Prove:
Passing point A is the extension line of AG‖BC intersecting DF at G?
AF/FB=AG/BD？ ,? BD/DC=BD/DC？ ,? CE/EA=DC/AG？
Multiply by three formulas:?
AF/FB×BD/DC×CE/EA = AG/BD×BD/DC×DC/AG = 1？
Its inverse theorem also holds:
If there are three points F, D and E on the side of AB, BC and CA or their extension lines, and AF/FB×BD/DC×CE/EA= 1, then the three points F, D and E are * * * lines.
Using this inverse theorem, we can judge the trisection line. ?
In addition, many people will think that writing this formula is too complicated to remember without reading it. Here are some ways to help you write?
1.ABC is three vertices and DEF is three points?
(AF/FB)×(BD/DC)×(CE/EA)= 1？
(Up/Down) * (Up/Down) = 1?
2. In order to explain the problem and impress everyone, we assume that A, B, C, D, E and F in figure 1 are six tourist attractions, which are connected by roads.
We flew over these scenic spots by helicopter and then chose any one to land. We transferred to the bus, went to every scenic spot along the expressway, and finally returned to the starting point, where the helicopter stopped waiting for us to go back. ?
We don't have to consider how to take the shortest route, we just need to "visit" all the scenic spots. A scenic spot that only "passes by" without stopping to watch is not a "tour". ?
For example, when the helicopter landed at point A, we set off from point A, "visited" the scenic spots represented by the other five letters, and finally returned to the starting point A ...?
Another requirement is that three scenic spots on the same straight line must swim continuously before they can be changed to other scenic spots on the straight line. ?
There are four kinds of travel plans from point A, which are explained one by one below.
Scheme? ①? From A to F, through B (non-stop), back to B (non-stop), to D (non-stop), through B (non-stop) to C (non-stop), to E (non-stop), and finally from E to C (non-stop) to the starting point A? ? According to this scheme, you can write the relationship:? (AF:FB)*(BD:DC)*(CE:EA)= 1 .？
What are the travel plans from point A?
Scheme? ②? It can be abbreviated as: A→B→F→D→E→C→A, from which the following formula can be written:
(AB:BF)*(FD:DE)*(EC:CA)= 1 .
Starting from a, you can also go in the direction of "C", so there are:
Scheme? ③? ——? A→C→E→D→F→B→A，
From this, the formula can be written as:? (AC:CE)*(ED:DF)*(FB:BA)= 1 .？
Starting with a, there is one last plan:
Scheme? ④? ——? A→E→C→D→B→F→A，
Write the formula from this:? (AE:EC)*(CD:DB)*(BF:FA)= 1 .？
Our helicopter can also choose to land at any point of B, C, D, E and F, so there are some other formulas in Figure 1. ?
Although figure 1 lists many formulas, not all of them can be written. ?
It is worth noting that some formulas contain four factors instead of three in Menelaus theorem. When the helicopter lands at point B, there are four factors. At points C and F, there are three formulas and four formulas. In Formula 4, some scenic spots will be visited twice. ?
Can we say that we have a deeper understanding of Menelaus theorem now? Those complicated multiplication and division relationships are written incorrectly and can't be remembered. ?
Menelius theorem? -? Inverse principle
The inverse theorem also holds: if there are three points F, D and E on the sides of AB, BC and CA or their extension lines, and AF/FB×BD/DC×CE/EA= 1, then the three points F, D and E are * * * lines.
Using this inverse theorem, we can judge the trisection line.