That is, there are points P, Q and R on the three sides BC, CA and AB of △ABC or their extension lines, then the necessary and sufficient conditions for P, Q and R to be collinear are as follows.
Necessary and sufficient conditions BP/PCXCQ/QAXAR/RB= 1.
Folding proof 1:
Passing point A is the extension line of AG∨BC passing through 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
Proof 2: Make a straight line AM∨FD at point M through BC extension line, then
CE/EA=CD/DM, AF/FB=MD/DB, so
(DB/DC)X(CE/EA)×(AF/FB)=(DB/DC)X(CD/DM)X(MD/DB)= 1
Another certificate:
Connecting CF and AD is based on the property that the area ratio of two triangles at the same height is equal to the ratio of the bottom.
Af: FB = s △ ADF: s △ BDF ………… ( 1)，BD:DC = s△BDF:s△CDF…………(2)，CE: EA = s △ CDE: s △ ade = s △ FEC: s △ FEA。 ):(S△ade+S△FEA)= S△CDF:S△ADF…………………………(3)( 1)×(2)×(3)(AF:FB)×(BD:DC)×(CE:EA)=(S)。
The inverse theorem of folding it also holds: if there are three points F, D and E on the sides of AB, BC and CA or their extension lines, AF/FB×BD/DC×CE/EA= 1, then F, D and E are collinear. Using this inverse theorem, we can judge that three points are collinear.
(1) This theorem can be proved by Menelaus Theorem (referred to as Mayer Theorem for short): ∫△ADC is cut by a straight line ∴ (CB/BD) * (Do/OA) * (AE/EC) =1① ∫△ Abd is cut by a straight line ∴. ×(af/ea)=S△AOB/S△AOC ③ Similarly CE/EA=S△BOC/ S△AOB ④, AF/FB = S △ AOC/S △ BOC σ ③× ④× ⑤ (BD/DC) * (CE/EA) * (AF/FB). ① Using the inverse theorem of Seva's theorem, it is proved that the three high lines of a triangle must intersect at one point: let the heights of the three sides of △ABC be AE, BF and CD respectively, and the vertical feet be D, E and F respectively. According to the inverse theorem of Seva's theorem, because (ad: db) * (be: EC) * (cf: fa) = [(CD * cot ∠ BAC. ② The three midlines of the triangle intersect at one point (center of gravity): as shown in the right figure, D and E are the midpoints of BC and AC of △ABC respectively, connecting AD and BE intersect at O point, connecting CO and extending the intersection of AB at F point. It is proved that AF=FB: ∵BD=DC, CE = EA ∴ BD/DC = 65438. Ce/EA =1(AF/FB) * (BD/DC) * (CE/EA) =1∴ AF/FB =1∴ AF = FB, ∴CF is the center line of △ABC. Therefore, the necessary and sufficient condition for the intersection of AL, BM and CN is λμν= 1. (Note that it is different from Menelaus theorem, where λμν=- 1)3 Edit 1. The necessary and sufficient conditions for the angular shape of Seva's theorem to BE that AD, BE and CF intersect at one point are: (sin ∠ bad/sin ∠ DAC) * (sin ∠ ACF/sin ∠ FCB) * (sin ∠ CBE/sin ∠ EBA) =/kloc. The necessary and sufficient condition for CF to intersect at a point is: (ab/BC )× (CD/de )× (ef/fa) =1,which can be easily proved by the angle form of Seva theorem and sine theorem and the relationship between chord length and the angle of the circle.