Lawson’s Geometrical Theorems. 449 
triangles EA: ED:: AB = DC: AL, 
and HA: AD = BC :: BF: LF’; conse- 
quently the rectangle EDC + rectangle 
CBF — EA x (AL + LF) = EA. AP. 
Which is Proposition 47. 
Draw FK parallel to DE, then FE : 
CE :: KF = CD = AB: AL, and FE: 
CF: : BF: “oe therefore FE x (AL + 
LF) = x AF = ECD + CFB 
(DKA). or: is Proposition 48. 
Also AF : AB + BF :: AB+ BF: 
AL+BL+ BL+ LF ~ AF + 2BL. 
Which is Proposition 53. 
And AF — BF: BF — LF:: BF: 
LF, because AF : BF: : BF : LF. Pro- 
position the 5ath. 
Diacram XI. 
if CB — AE = EB = CE, and CF 
= FB, then the angle FCB = FBC, and 
47 
53 
55 
NS Fe 
then I say that the square of HF is to the square of the 
tangent HC as FI to IG. 
Prop. LVH. If one side AC of an equilateral triangle 
ABC be produced to Eso that CE may be equal to AC, and 
from A a perpendicular to AC raised, and from E a line 
drawn through the vertex B to meet the perpendicular in 
D; then I say that BD-is equal to the radius of the circle 
which circumseribes the triangle. 
3L 
