% 
336 HERON OF ALEXANDRIA 
Also 
A ABC: A ABB = A ADC: A EDO. 
But (since the area of the triangle DEF is given) A EDO is 
given, as well as A ABC. Therefore A ABD x A A DC is given. 
Therefore, if AH he perpendicular to BG, 
AH 2 , BD. DC is given ; 
therefore BD. DC is given, and, since BG is given, D is given 
in position (we have to apply to BG a rectangle equal to 
BD. DC and falling short by a square). 
As an example Heron takes AB =13, BG =14, G A =15, 
A DEF = 24. A ABC is then 84, and AH =12. 
Thus AEDG = 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 
Book III. Divisions of figures. 
This book has much in common with Euclid’s book On divi 
sions {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 » 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 v 
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, GED 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 
each of the three ratios which will 
result in the triangle DEF having a 
given area. 
Join AD. 
Since BD-.GD = GE-.EA, 
BG: CD = GA : AE, 
and A ABC: AADG=AADG:AADE.