ON DIVISIONS OF FIGURES 
429 
figure bounded by an arc of a circle and by two straight lines 
which form a given angle. 
Let A В EC be the given figure, D the middle point of BG, 
and DE perpendicular to BG. Join AD. 
Then the broken line ADE clearly divides the figure into 
two equal parts. Join AE, and draw 
DF parallel to it meeting BA in F. 
Join FE. 
The triangles AFE, ADE are then 8 
equal, being in the same parallels. 
Add to each the area AEG. 
Therefore the area AFEG is equal to the area AD EG, and 
therefore to half the area of the given figure. 
(3) Proposition 29 : To draw in a given circle two parallel 
chords cutting off a certain fraction (in/n) of the circle. 
(The fraction m/n must be 
such that we can, by plane 
methods, draw a chord cutting off 
m/n of the circumference of 
the circle; Euclid takes the case 
where m/n — -|.) 
Suppose that the arc AD В is 
m/n of the circumference of the 
circle. Join A, В to the centre 0. 
Draw OC parallel to A В and join 
A G, BG. From D, the middle point 
of the arc AB, draw the chord DE parallel to BG. Then ghall 
BG, DE cut off m/n of the area of the circle. 
Since AB, OC are parallel, 
Д АО В = A ACB. 
Add to each the segment AD В; 
therefore 
(sector ADBO) = figure bounded by AC, CB and arc ADB 
= (segmt. ABC) — (segmt. BFG). 
Since BG, DE are parallel, (arc DB) = (arc GE);