A history of Greek mathematics - Wilbourhall.org

424 PAPPUS OF ALEXANDRIA
And, since AFB.EFG are both right angles, lBFG = lAFE.
But, since A, E, G, F are concyclic, LAFE — A AGE.
Therefore
IBDG = I AGE;
and the right-angled triangles DBG, GAE are similar.
Therefore AG : AE = BD: GB,
or
AG.GB = AE.DB.
In Apollonius G and the corresponding point G' on BA
produced which is obtained by drawing F'G' perpendicular to
ED (where DE meets the circle again in F') are the foci
of a central conic (in this case a hyperbola), and DE is any
tangent to the conic ; the rectangle AE .
to the square on half the conjugate axis.
BD
is of course equal
(77) The Lemmas to the Conies of Apollonius (pp. 918-1004)
do not call for any extended notice. There are a large number
of propositions in geometrical algebra of the usual kind,
relating to the segments of a straight line marked by a number
of points on it ;
propositions about lines divided into proportional
segments and about similar figures ; two propositions
relating to the construction of a hyperbola (Props. 204, 205)
and a proposition (208) proving that two hyperbolas with the
same asymptotes do not meet one another. There are also
two propositions (221, 222) equivalent to an obvious trigonometrical
formula. Let ABGD be a rectangle, and let any
straight line through A meet DC produced in E and BG
(produced if necessary) in F.
Then EA .
AF = ED . DC + CB .
BF.