Lawson's Geometrical Theorems. 487 
Hence we have a ready way of finding 
two such points 3, y in the perpendicular 
that 3y7 may = the sum of the squares of 
the tangents from these points to the cir- 
cle. Let 3, yin Diagram 4, be two such 
points, and from y, let a line be drawn 
meeting the circle in any two points E’ 
and H’, then 3H’, aK’ being drawn meeting 
the circle againin hand D’, because 3H’. 
dh = SF’ . 3 = 9D’. 3E’; therefore F, h, H’, y 
are in a circle, as also F', D’, E’, v; therefore 
the 2 FyH = 2D’F = 3dbP, therefore 3, h, 
D’, F are in a circle, and 73D’“h = 3Fh = 
3H’y, also FD’y = F2h; consequently the 
three angles 2D’h, 3D’F’, yDF at the point 
D’ being respectively equal to. 3Fh, shF, 
Fh, those of the triangle Feb, their sum 
must be equal to two right angles, and 
tiple of the semidiameter of the circle by the number of the 
sides of the figure, 
Prop. XXXIX. Let there be any number of right 
lines intersecting in a point, and making all the angles about 
the point equal, and let any circle pass through the same 
point; Isay the circumference thereof will be divided by 
the intersecting lines into as many equal parts as there are 
lines. 
Prop. XL. If there be two triangles ABC, DEF, 
~ which have one angle A in one equal to one angle D in the 
