Join AE, AQ, and QB ; then, because BD is bisected
in E, the triangle ABE is equal to half the triangle
ABD ; and since GQ=iGE, the triangle AGQ is equal
to half the triangle AGE, and the triangle BGQ is equal
to half the triangle BGE ; hence the ti-iangle ABQ=i
the triangle ABE, and, consequently, the triangle ABQ
=i the triangle ABD.
Now, suppose BE and DE were each bisected, and
from the points of bisection lines were drawn parallel to
DA, these lines would evidently bisect BG and GA
respectively. Then, by drawing lines from the points in
which those parallel lines would cut the curve to the
extremities of the chords BQ and QA, the sum of the
triangles formed within the spaces BQ and QA will be
equal to J the triangle ABQ, or -jlg-th of the triangle ABD.
By bisecting the halves of BE and ED, and drawing
lines as before parallel to AD, and joining the points of
their intersections with the curve to the extremities of
the chords, a series of triangles would be formed in the
remaining spaces, the sum of which would be equal -to
^th of the triangle immediately preceding it, or equal to
■^th of the triangle ABD.
Continue this bisection indefinitely, and we shall have
the triangles filling the whole parabola, in which case
the area of the parabola would ultimately be equal to



ON THE ELLIPSE AND HYPERBOLA.
45



the sum of the areas of all the triangles thus formed
within it.
Put a=;the area of the A ABD.
