4 
Lawson's Geometrical Theorems. AA47 
similar, consequently EB =FB:CE:: 5 
AF: CF. Whiehis Proposition 58.: 
Having taken CFL = CBF, it will be 
CB: CF: : CF: CL, andthe angle 
CLF being = CFB, ALF must be = 
AFC, and the triangles ALF, AFC 
similar ; therefore AC: AF ::AF : AL. 59° 
Which is Proposition 59, 
If the cireumscribng circle be rion 
about the triangle ACB, and HL be 
drawn. parallel to AB to eut it m H and 
L’, and CH, Cl be joined cutting the 
base in M and N, and making equal 
angles ACH, BCL with the sides; then 
the rectangle AMB = CMH, and CNL’ 
‘= ANB, but CM: CN : : MH: NU’ 
therefore CM*: CN*:: CM. MH; CN. 
NL’:: AMB: ANB. Which is Propo- 45 
sition 45. 
[4 
angle, sois the said sum, to the sum of the hypothenuse 
and twice the perpendicolar from the right angle. 
Prov. LIV. If aright line AD be any-ways cut in B, 
and from thence a perpendicular BE erected equal to a 
mean proportional between the whole AD and the part AB, 
aud a circle be drawn through the points A, D, E, and 
from A perpendicular be erected to meet the circumference 
in F; them Lsay that AF, AB, BE, AD are four continual 
proportionals, 
