428 
EUCLID 
internal point; and suppose the problem solved, i. e. GH 
drawn through D in such a way that A GBH = ^ • A ABC. 
Therefore GB. BH = —. AB. BG. (This is assumed by 
Euclid.) 
Now suppose that the unknown quantity is GB = x, say. 
Draw DE parallel to BG; then BE, EB are given. 
Now BH.DE = GB:GE=x: {x-BE), 
or 
therefore 
BH = 
x. 
X — 
GB. BH — x* 
DE 
He'' 
DE 
’ x — BE ‘ 
A 
And, by hypothesis, GB. BH = — . AB . BG; 
n 
,, 2 m AB .BG . 
therefore x* = -. - -gg- (x - BE), 
7 m AB. BG 
or, it Ic — jjj-j—, we have to solve the equation 
x 2 = k (x-BE),' 
or kx — x 2 = k . BE. 
This is exactly what Euclid does; he first finds F on BA 
such that BF. DE • AB. BG (the length of BF is deter 
mined by applying to DE a rectangle equal to AB. BG, 
Eucl. I. 45), that is, he finds BF equal to k. Then he gives 
the geometrical solution of the equation kx — x 2 = k. BE in the 
form ‘ apply to the straight line BF a rectangle equal to 
BF .BE and deficient by a square’; that is to say, he deter 
mines G so that BG . GF = BF .BE. We have then only 
to join GD and produce it to H; and GH cuts off the required 
triangle, 
(The problem is subject to a Siopur/xos which Euclid does 
not give, but which is easily supplied.) 
(2) Proposition 28 : To divide into two equal parts a given