DIVISIONS OF FIGURES 
337 
should apparently have 8|, since DC is immediately stated to 
be 5^ (not 6). That is, in solving the equation 
¡r 2 — 14# + 46f = 0, 
which gives x = 7 + V(2-|), Heron apparently substituted 2\ or 
| for 2|-, thereby obtaining 1^ 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 + n), dm. 
Hence 
A AFE /A ABC = ■ *™^ 2 =ABDF/A ABC = ACDE/AABC, 
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 (1) 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 §*rea of a 
circle into parts which have a given ratio by means of an 
inner circle with the same centre. For the problems begin 
ning 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 
HI. 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, (h) 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, (h) in the case where it is not; the 
last case (h) is reduced (like III. 10) to the ‘ cutting-off of an 
1 Pappus, viii, pp. 1034-8. Cf. pp. 430-2 post. 
I 
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