414 
PAPPUS OF ALEXANDRIA 
I need only draw two figures by way of illustration. 
In the first figure (Prop. 83), ABC, DEF being the semi 
circles, BEKG is any straight line through C cutting both; 
FG is made equal to AD; AB is joined; GH is drawn per 
pendicular 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'GH' is 
a tangent, then B'M' = M'H' (Props. 83, 84). 
In the second figure (Prop. 91) D is the centre of the 
semicircle ABC and is also the extremity of the diameter 
of the semicircle DEF. If BEGF be any straight line through 
Fig. 2, 
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 ABC on the diameter AC 
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 at E and meeting the semicircle 
ABC in B, BE shall be equal to AD.