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