122 ESSAYS aNnp OBSERVATIONS 
the triangle DFM equal to the fector DFG: 
But the other way points out the method 
of calculation. 
PROP. IV. Prosu. 2. Fig, 4. 
Let there be a femicircle whofe diameter 
is AB and centre C, and let D be a point 
in the diameter: Granting the quadra- 
ture of the circle, it is required to draw 
a line DE meeting the circle in F, fo 
that the femicircle may be to the fector 
BDE in a given ratio, fuppofe that of 
p to 4. 
Suppose the problem folved. Let the 
f{emicircle be to the fector BCF as pis tog; 
join DF; draw CG parallel to DF, meet- 
ing FG, a tangent to the circle at F,inG, 
and join DG meeting the circle in H; let 
CG meet the circle in K, and join DK, 
FK, CH. 
Because the femicircle is to the fector 
BDE as ? tog, that is, as the femicircle 
to the fector BCF, the fector BDE will be 
equal to the fe@tor BCF: But the fector — 
BCF 
