AM, BN, CP concurrent <=> MB/MC × NC/NA × PA/PB = -1 |

An elementary proof results from noting that ratio NC/NA equals ratio of distances
from C and A to line BN.

Triangles BQA and BQC, with same base BQ, have then their areas in the same ratio.

That is : NC/NA = (BQC)/(BQA).

Similarily MB/MC = (AQB)/(AQC) and PA/PB = (CQA)/(CQB)

Hence MB/MC × NC/NA × PA/PB = 1 in absolute value,
and equals -1 taking the signs of ratios.

Conversly let P' the intersection point of AB and CQ, we get MB/MC × NC/NA × P'A/P'B = -1 from the direct theorem. Hence P'A/P'B = PA/PB then P and P' are the same point.

Application : Gergonne point, Nagel point

Given a triangle, its incircle touching the sides in M, N, P.

AP = AN, BP = BM and CM = CN, hence MB/MC × NC/NA × PA/PB = -1

Lines AM, BN and CP then intersect in one point : the Gergonne point.

Similarily with contact points of excircles.

From the length of common tangents we easily deduce AQ = BP, AR = PC and BR = QC

For instance AU = AV that is AB + BP = AC + CP = AC + (BC - BP) hence BP = (AC + BC - AB)/2

Similarily BA + AQ = BC + CQ = BC + (AC - AQ) hence AQ = (BC + AC - BA)/2, that is BP = AQ

Immediately results that AP, BQ and CR intersect in one point : the Nagel point.

Angle variant

Area of BQA = 1/2 BQ.BA.sin(QBA) and similarily for other triangles.

Finally results :

AM, BN, CP concurrent <=>
sin(NBA)/sin(NBC) × sin(PCB)/sin(PCA) × sin(MAC)/sin(MAB) = -1 |

Application : symmedians, Lemoine point

Let AM a median and AI the angle bissector of A, AU the symetric of the median from the angle bissector.
Lines AM and AU are said "isogonal" : angles BAU = MAC and also BAM = UAC.

AU is then called a "symmedian".

sin(UAB)/sin(UAC) = 1 / ( sin(MAB)/sin(MAC) ) hence

Lines AM... concurrent ⇔ lines AU... also concurrent.

Three lines are concurrent if and only if their isogonal are concurrent |

Specifically the symmedians intersect in one point : the Lemoine point.