A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

DIVISIONS OF FIGURES 337
should apparently have 8 J, since DC is immediately stated
be 5J (not 6). That is, in solving the equation
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.
Let m : 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
mn
AAFE/AABC =
'
-aBDF/AABG = ACDE/AABG,
(m + ny
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 £
to
338 HERON OF ALEXANDRIA
area'. These propositions are ingenious and interesting.
III. 11 shall be given as a specimen.
Given any quadrilateral ABCD and a point E on the side
AD, to draw through E a straight line EF which shall cut
the quadrilateral into two parts in
the ratio of AE to ED. (We omit
the analysis.) Draw CG parallel
to DA to meet AB produced in G.
Join BE, and draw GH parallel
to BE meeting BC in H.
Join CE, EH, EG.
Then AGBE = AHBE and, adding AABE to each, we have
AAGE = (quadrilateral ABHE).
Therefore (quadr. ABHE) : ACED = A GAE: ACED
= AE:ED.
But (quadr. ABHE) and ACED are parts of the quadrilateral,
and they leave over only the triangle EHC.
We have
therefore only to divide A EHC in the same ratio AE-.ED by
the straight line EF. This is done by dividing HC at F in
the ratio AE: ED and joining EF.
The next proposition (III. 12) is easily reduced to this.
If AE : ED is not equal to the given ratio, let F divide AD
in the given ratio, and through F
draw FG dividing the quadrilateral
in the given ratio (III. 11).
Join EG, and draw FH parallel
to EG. Let FH meet BC in H,
and join EH.
Then is EH the required straight
line through E dividing the quadrilateral
in the given ratio.
For AFGE = AHGE. Add to each (quadr. GEDC).
Therefore (quadr. CGFD) = (quadr. CHED).
Therefore EH divides the quadrilateral
in the given ratio,
just as FG does.
The case (III. 13) where E is not on a side of the quadrilateral
[(2) above] takes two different forms according as the