AAA A Demonstration of 
the sides, the sum of their double Areas, 
must be equal to the product of the sum 
of their perpendiculars and the side of 
the polygon. Therefore the radius x 
number of sides = the sum of the perpen- 
38 diculars. Which is Proposition 38. 
And since equal arches of the same 
circle subtend equal angles as well at the ' 
39 circumference as centre, therefore the 
39th. Proposition is manifest. 
meee oy FD B, 
Because AD + DB = AB, therefore 
AD + BC =CD + AB, and AB + AD 
+ BC =2AB + CD, consequently (AB 
+ AD) .CB + CB* = 2ABC + BCD. 
Also AB + CD + AD=2AD + BC, 
therefore (AB + AD). CD + CD? = 
44 2ADC + BCD. Which is Proposi- 
tion 44. 
the rectangles under the segments of the sides about the 
right angle made by this inscription. 
Pror. XLIX. If from the same point C two tangents 
be drawn to a semicircle whose diameter is AB, and if the 
extremes of the diameter and the points of contact be 
joined, either cross-ways by two lines intersecting in F, or 
other-ways by two lines intersecting in H; then I say that 
CF or HC produced to meet the diamater AB will be per- 
pendicular to the-same. 
