f 299 '] 
PROPOSITION III. 
fhe fquare of the fide of an equilateral triangle infer ibeddn 
a circle is equal to a reSi angle under 'the > diameter of the 
circle^ and a perpendicular let fall from -any angle -of the 
triangle upon the oppGfitefide> Fig, 4. , 
THE Jtwo triangles adc, aec, are .equi-angular and 
fimilar, the angles acd,, A.Ec,,'heing both right, and that 
at A common; therefore ad : ac ac : ae, and ac^ == 
AD X AE. 
The fquare of one lide of the triangle being com- 
pleated fo as to include the triangle, 1 fay, that part 
of the tide of the fquare that falls within the circle 
is equal to the radius ; and the other part, lying with- 
out the circle, is equal to the radius minus twice the 
portion lying between the lide of the fquare, and the 
circumference of the circle ; or is equal to that part of 
the radius that lies between the centre and the lide of 
the fquare minus the remainder of the radius ; that is cl 
(a) And univerfally, a perpendicular being drawn from any angle of a right- 
lined triangle to the oppofite fide, the reftangle under the two fides which contain 
that angle, is equal to the reftangle under the perpendicular and the diameter of 
the circumferibed circle. (See tab. viii. fig. 5.) 
From A, one angle of the triangle bac, draw ae perpendicular to the fide 
BC. Round the triangle bac deferibe a circle, and draw the diameter ad. I fay, 
the reftangles ac x ab, ae x ad, are equal. Join db. The angle dea is a 
right angle. Therefore it is equal to the right angle cea. The angles at the 
circumference ace, adb, are equal, becaufe they ftand upon the fame arc ab. 
Therefore the two triangles ace, adb, are fimilar, and ac : AE ~ ad : ab. 
Xfierefore ac x ab:=; ae X ad, Z>» s, HOASLEy, 
IS 
