A history of Greek mathematics - Wilbourhall.org

414 PAPPUS OF ALEXANDRIA
I need only draw two figures by way of illustration.
In the first figure (Prop. 83), ABC, DEF being the semicircles,
BEKC is any straight line through C cutting both;
FG is made equal to AD; AB is joined; GH is drawn perpendicular
to BK produced. It is required to prove that
BE = KH. (This is obvious when from L, the centre of the
semicircle DEF, LM is drawn perpendicular to BK.) If E, K
coincide in the point M' of the semicircle so that B'CH' is
a tangent, then B'M' = M'H' (Props. 83, 84).
In the second figure (Prop. 91) D is the centre of the
semicircle ABG and is also the extremity of the diameter
of the semicircle DEF. If BEGF be any straight line through
F cutting both semicircles, BE — EG. This is clear, since DE
is perpendicular to BG.
The only problem of any difficulty in this section is Prop.
85 (p. 796). Given a semicircle ABG on the diameter AG
and a point D on the
diameter, to draw a semicircle passing
through D and having its diameter along DC such that, if
CEB be drawn touching it
ABC in B, BE shall be equal to AD.
at E and meeting the semicircle