Llandaff –
Wednesday evening. Nov 30

My dear Father.

I send you a geometrical theorem which will I hope prove to be correct. I give the figure on another sheet. Let 3 planes EDB. EDC. BDC pass through one point D. and let 2 other planes EBC. FGH make triangular sections with these 3 planes. The corresponding sides of these triangles EBC & FGH not being parallel. Then if the corresponding sides EC & FH. BE. & GF BC &GH be produced to meet. The points of intersection – M. A & K shall all lie on one straight line For because BE and FG are in one plane EBD and not parallel they intersect in A.

and sin also since BC. GH are in one plane BDC and not parallel they intersect in K.

Therefore the planes. BEC. and GFH produced have for their common section the straight line AK.

Again. BC EC GH FH are in one plane BCD ECD and not parallel they intersect in K M. which is necessarily a point on AK the intersection of the planes EBC. FGH the 3 pts of intersection are one on one straight line

And since that is true however the triangles EBC. FGH are drawn the geometrical theorem follows. If 3 lines passing

Now suppose BCD. to lie in the plane of the paper.

The theorem is true however near E. approaches the plane of the paper. it is true in in the limit when E is indefinitely near the plane of the paper. Then AKM are also in the plane of the paper, and we have the theorem in for a plane

If 2 straight lines It is The general ennunciation [sic] does for both cases. If 3. straight lines passing through the point D. pass each through 2 of the angles of a triangle. Then if the corresponding sides of the triangles be produced to intersect, the 3 points of intersection lie all on one straight line.

Your affect son
– Charles. –

[envelope]
H Fox Talbot Esq
Lacock Abbey
Chippenham