Mathematical Problem. 109 
To the Editor of the Monthly Magazine. 
“ Brevis esse laboro, obscurusyfio. 7, 
‘Sir: eit 
} AVING observed, in, your; last, number, the 
solution of an ingenious, and, to me,\new 
Geometric Theorem, I could not avoid considering 
some of the steps of the demonstration. rather 
obscure, and the conclusion, geometrically, un- 
satisfactory. Under this impression, I enclose,a 
proof, substantially the same as the former, but 
which will, I think, atone for its greater length, by 
increased perspicuity and strictness. $ 
[We apprehend that the motto affixed to this commu- 
nication is applied to our former very clever correspon- 
dent, Mr. Davies: perhaps he may be induced to remoye 
any imputation of this kind, without referring -to. the 
pedantries, or the Porisms, of almost unknown authors,— 
Epir. | 
Uron either pair of opposite sides of a trape- 
zium, as BC, DA, let the triangles BEC, AFD be 
constructed, each having its vertex any where in 
the other’s base: then, if the sides of the triangles 
intersect in G and O, and the diagonals of the tra- 
pezium intersect each other in K, the points, G,K,0O, 
are in the same straight line. 
1. Let BC be parallel to AD :— 
Join GK, and produce it both ways, to cut CB 
and AD produced ;—then, if GK produced does not 
pass through O, let it cut FD in H’, and CE in H. 
Therefore, because LC is parallel to AM, 
“ LB: ME:: BG: GE, 
and ME:LC :: EH: HC. 
.. LB: LC :: BG. EH: GE.HC. 
Again, MD:LF:: HD: HF, 
and LF:MA:: FG: GA. 
. MD: MA:: FG.H'D: GA. HF. 
Also, LB: LK :: MD: MK, - 
and LK : LC :: MK: MA. 
LB: LC :: MD: MA. 
Hence, BG. EH : GE. HC =: FG. HD : GA. HF; 
but, from the similar triangles, BGF, AGE, 
BG:GE:: FG: GA, 
», EH: HE : HD: HF. 
Or, EO+OH : OC—OH :: DO+0OH’: OF—OH’. 
-7+ comp EO+0C : OC—OH :: DO+OF : OF—OHW’ (2) 
But, from similar triangles, EO : OC: DO: OF, 
“. EO+0C: OC ::; DO+OF : OF. 
Invert? OC :EO+O0C :: OF: DO+OF (8) 
-« By comparing (2) and (8), OC : OC—OH :: OF: OF— OH". 
ny Conve: OC: OH :: OF: OH’. 
22894 Permut? OC: OF :: OH: OH’. 
.". (Buelid, vi. 2), HH’ is parallel to CF; but HH’ is in the same straight line 
with GK, .*. also, GK is parallel to CF; and, if GK and CF be produced ever 
so far; they will not meet. But, GK being produced, does meet CF, produced in 
LL; which is absurd: .*. H’H, or KH, is not in the same straight line with GK; 
and inthe same manner it may be shewa, that no other than ‘KO can be in the 
same straight line with GK. Wherefore, the points, G,K,O, are in the same 
straight line, Q. E. D. 
“The lines, with the exception of the parallelism of BC, AD, being arbitrary, we haye. 
merely to conceive the figure laterally projected upon an oblique plane, when the repre- , 
sefitation will be a trapezium perfectly unlimited in the conditions of its structure, and 
having all the coincidences stated in the theorem.” 
