246 FLEMENTARY DEMONSTRATION OF THE 
Let ABCD be a rec- FIG.1 
tangle, and AC the dia- 
gonal passing through eo 
A; draw EAF perpen- ” 
A 
dicular to AC, and let 
fall the perpendiculars — fort 
BF, BH, DG, DE: then x 
shall ABCD, AFBHand =» c 
AEDG be similar rectangles ; and if, in each of these, the 
equivalent of the pressures represented in direction and quan- 
tity by the sides, be represented in direction by the diagonal, 
it shall be represented by the same in quantity also. For if 
the forces AH and AF be equivalent to m AB; AE and AG 
shall be equivalent to m AD; and AB and AD to m AC; or 
m AB and mAD to m AC: that is, the forces AH, AF, AE, 
and AG, will be equivalent to m AC: But AE and AF are 
equal and opposite : hence the forces AH and AG are equi- 
valent to m= AC. But AH and AG are equivalent to AC; 
therefore m = 1. 
1. Now let ABDC be any square ; and let the sides AB and 
CD be produced indefinitely towards B and D; draw the dia- 
gonal AD; in CD produced, take DF equal to AD ; join AF; 
take FH equal to AF; join AH, and so on. 
the rectangles ACFE, ACHG, &c. 
It is obvious, that AD, AF, AH, &c. bisect the angles 
BAC, BAD, BAF, &c. respectively. Hence the resultant of 
AB 
And complete 
