Lawson's Geometrical Theorems. 441 
a tangent, the TBO in the segment is 
= FTG, consequently THT’ = TFG = 
TFG, therefore F'T’ being = FT, and 
the angle TF'I’ double THT’ F must be 
the centre of the circle TOT’H, conse- 
quently the diameter HO passes through 
F. Also since the 2OHT’ = OTT’ = 
OBA = the complement of PAH, the 
ZHPA isright. Which is the forty-ninth 49 
Proposition. 
Moreover FE’? : FT? = FE.Fh’: : FE’: 
Fl: : (by Prop. 15.) E’g’ : g’h’, therefore 
FE”? : FT” :: H’g’: g’h’. Which is the 56 
JSifty-sixth Proposition. 
Diacram VI. 
Having described a circle about the 
triangle BAC, and produced AD till it 
cuts in G, draw DF’, ED parallel to AB, 
Prop. XLV. If from the vertical angle of any triangle 
two lines be drawn to make equal angles with the sides 
containing it, and to cut the base; then I say that the 
square of one side is to the square of the other side asthe 
rectangle under the segments of the base contiguous to the 
first side is tothe rectangle under the segments contiguous 
to the other side. 
Prop. XLVI... If in AB the diameter of a semicircle 
any point C be taken, and from thence any line as CD 
3K 
