Consider the triangles AGC and HDM which are in perspective from K .
Since AG \cap HD = S and CG \cap MD = R , Desargues' theorem gives AC \cap HM = T as required. And, by symmetry, BD \cap GN = T .
Diagram 1


Consider the triangles GKC and BTH . Since KG \cap BT = D , GC \cap BH = R and CK \cap TH = M are collinear,
the converse of Desargues' theorem implies that the triangles are in perspective from MC \cap GB = L so KT passes through L as required.
Diagram 2


As I have not found a proof like the above, I refer to the proof in and give a projective version.
First we prove the special case that, if GHT are collinear, then so are IJT. Let X = AH \cap BG and consider the following projections: RGIC \rightarrow ^{H} BGXL \rightarrow ^{T} AHXK \rightarrow ^{G} SHJD .
The composite is equivalent to the projection by RGC \rightarrow ^{T} SHD so, since it is known that a projectivity is defined by the images of three of the points, IJT are collinear in this case.

For the general case, let S' = HA \cap ST, H' = RB \cap GT, I' = RC \cap AH' , D' = SH' \cap AB, J' = GD' \cap SH' , so we know that I'J'T are collinear.
Project S'HIA \rightarrow ^{R} S''H'I'A \rightarrow ^{T} SH'J'D' \rightarrow ^{B} SHJD .
The composite projectivity is the projection from S'HA to SHD using T and must therefore take I to J.
Diagram 3


To show that E = RK \cap SL is on AB , look at triangles SHL and RMK that are in perspective from T .
Corresponding sides intersect in D = SH \cap RM , C = MK \cap HL and E = SL \cap RK which are therefore collinear.
Diagram 4

The point E plays an important role as will be seen from the following.
Let S' = AK \cap RT and R' = BL \cap RT , then ACET \rightarrow ^{R} AIKS'\rightarrow ^{T} BJLR' \rightarrow ^{S} BDET ,
which proves that the cross ratios ACET and BDET are the same.
Let G''= AR \cap CS , H''= BS \cap DS and F = HG \cap H''G'' , then F lies on AB and E = GH'' \cap HG'' .
In the Euclidean case, the parallelograms AG''CG and BHDH'' are similar and the centres of similarity are E and F .