A history of Greek mathematics - Wilbourhall.org

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