338 
HERON OF ALEXANDRIA 
area’. These propositions are ingenious and interesting. 
III. 11 shall be given as a specimen. 
Given any quadrilateral ABGD 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 A A 1 to ED. (We omit 
the analysis.) Draw CD parallel 
to DA to meet AB produced in G. 
Join BE, and draw GH parallel 
to BE meeting BG in H. 
Join CE, EH, EG. 
Then A GBE— AHBE and, adding A ABE to each, we have 
A AGE — (quadrilateral ABHE). 
Therefore (quadr. ABHE): AC ED = A GAE: AGED 
= AE:ED, 
But (quadr. ABHE) and AGED are parts of the quadri 
lateral, and they leave over only the triangle EHG. We have 
therefore only to divide A EHG in the same ratio AE-.ED by 
the straight line EF. This is done by dividing HG 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 A 7 divide AD 
in the given ratio, and through F 
draw FG dividing the quadri 
lateral in the given ratio (III. 11). 
Join EG, and draw FH parallel 
to EG. Let FH meet BG in H, 
and join EH. 
Then is EH the required straight 
line through E dividing the quad 
rilateral in the given ratio. 
For A FGE = A HGE. 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 quadri 
lateral [(2) above] takes two different forms according as the