Lawson's Geometrical Theorems. 417 
AC inG, AH in H;; and bisect DH in 
n. Then because the triangles DGC, AGH 
are similar, DC : AH: : DG: GH; and 
because BFD, AFH are similar, BD = DC 
:AH:: FD: FH. Therefore by equality 
of ratio DG: GH: : FD: FH. Which is 3 
the third Proposition. 
Join EF, EG cutting BC in k and 1], 
through G draw Gi |jto EH cutting ED in 
o and EF in i; then by reason of the pa- 
rallel lines, AH : EH :: BD: Dk:: DC: 
Dl, and because BD = DC, .. Dk = DI; 
consequently ED bisects the 2 FEG, and 
EF: EG:: FD: DG. Which is the fifth 5 
Proposition. . 
Let fall Fm perp. to ED produced ; then 
LAWSON’S 
GEOMETRICAL THEOREMS. 
ad 
PROP. I. 
r a right line AB be bisected in E, and two points C and 
D taken therein such that AC : CB :: AD: DB; then 
I say the rectangle DC E = the rectangle ACB. 
The converse of this proposition is also true, which is 
this. 
If a right line A B be bisected in E, and two points C 
and D taken therein such that DCE = ACB; then I say 
AC: CB:: AD: DB. 
Pror. IL, If in AB the diameter of a circle two aes 
3G 
