A history of Greek mathematics - Wilbourhall.org

DIVISIONS OF FIGURES 337
should apparently have 8 J, since DC is immediately stated
be 5J (not 6). That is, in solving the equation
to
x 2 -14&' + 46§ = 0,
which gives x — 7 ± V(2±), Heron apparently substituted 2 J or
f for 2§, thereby obtaining \\ as an approximation to the
surd.
(The lemma assumed in this proposition is easily proved.
m Let : n be the ratio AF: FB = BD : DC = GE-.EA.
Then AF — mc/(m + n), FB = nc/(m + n), GE — mb/(m + n),
EA = nb/(m + ri), &c.
Hence
AAFE/AABC =
'
mn
(m + ny
-aBDF/AABG = ACDE/AABG,
and the triangles AFE, BDF, GDE are equal.
Pappus 1 has the proposition that the triangles A BG, DEF
have the same centre of gravity.)
Heron next shows how to divide a parallel-trapezium into
two parts in a given ratio by a straight line (l) through the
point of intersection of the non-parallel sides, (2) through a
given point on one of the parallel sides, (3) parallel to the
parallel sides, (4) through a point on one of the non-parallel
sides (III. 5-8). III. 9 shows how to divide the area of a
circle into parts which have a given ratio by means of an
inner circle with the same centre. For the problems beginning
with III. 10 Heron says that numerical calculation alone
no longer suffices, but geometrical methods must be applied.
Three problems are reduced to problems solved by Apollonius
in his treatise On cutting off an area. The first of these is
III. 10, to cut off from the angle of a triangle a given
proportion of the triangle by a straight line through a point
on the opposite side produced. III. 11. 12, 13 show how
to cut any quadrilateral into parts in a given ratio by a
straight line through a point (1) on a side (a) dividing the
side in the given ratio, (6) not so dividing it, (2) not on any
side, (a) in the case where the quadrilateral is a trapezium,
i.e. has two sides parallel, (b) in the case where it is not; the
last case (b) is reduced (like III. 10) to the ' cutting-off of an
1
Pappus, viii, pp. 1034-8. Cf. pp. 430-2 post.
1523 2 £