A history of Greek mathematics - Wilbourhall.org

336 HERON OF ALEXANDRIA
Book III. Divisions of figures.
This book has much in common with Euclid's book On divisions
(of figures), the problem being to divide various figures,
plane or solid, by a straight line or plane into parts having
a given ratio.
In III. 1-3 a triangle is divided into two parts
in a given ratio by a straight line (1)
passing through a vertex,
(2) parallel to a side, (3) through any point on a side.
III. 4 is worth description :
'
Given a triangle ABC, to cut
out of it a triangle DEF (where D, E, F are points on the
sides respectively) given in magnitude and such that the
triangles AEF, BFD, GET) may be equal in area.' Heron
assumes that, if D, E, F divide the sides so that
AF: FB = BD: DC = CE: EA,
the latter three triangles are equal in area.
He then has to find the value of
Also
each of the three ratios which will
result in
given area.
Join AD.
the triangle DEF having a
Since BD:CD = CE-.EA,
BC:CD= CA-.AE,
and AABC:AADC=AADC:AADE.
AABC: AABD = AADC: AEDC.
DEF is given) AEDC is
But (since the area of the triangle
Therefore AABD x A A DC is given.
Therefore, if AH be perpendicular to BC,
AH 2 .BD.DC is given;
given, as well as AABC.
therefore BD .
DC is given, and, since BC is given, D is given
in position (we have to apply to BC a rectangle equal to
BD .
DC and falling short by a square).
As an example Heron takes AB =13, BC =14, CA = 15,
ADEF = 24. AABC is then 84, and AH = 12.
Thus AEDC= 20, and AH 2 . BD. DC = 4 . 84
. 20 = 6720;
therefore BD .DC = 6720/144 or 46| (the text omits the §).
Therefore, says Heron, BD — 8 approximately. For 8 we