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branches on J. Let PQ be a line through A in « inside “ BAC 
(fig. 1) and y a plane through PQ, such that the branches AM and 
Bigot; 
AN depart to different sides. In y a branch departs from A on / 
in front of 8. This branch has the line of intersection of Band y for 
tangent, as otherwise 6 could not be the last plane with double point 
in A and branches on /. In y also departs a branch on ///, hence 
A is double point in y with the line of intersection of y and # as 
one of the tangents. The character of the curve in >, however, requires 
this line to carry a point of F* different from A, hence a contradiction 
is obtained *). 
Not all planes through RS show a double point at A, for the set 
of planes with 2 branches on J and the set of those with 2 on /// 
would be separated by a plane with 1 branch on / and /// each 
and these would be lines. In every plane through RS however, 2 
branches arrive at A from below «, hence a plane 8 through RS 
exists in which A is either ordinary point or cusp. Let us assume 
the first. Now let 8 turn round RS, then before «a is reached, A has 
become double point. The assumption that no plane through RS 
shows a cusp in A leads to an impossibility, for when $ turns, there 
cannot be a first plane with double point in A (as shown above) 
and that there cannot be a last position with an ordinary point in 
A is proved by the reasoning of comm. 1 p. 113 and 114. 
Hence there exists a plane through RS with a cusp in A. Let @ 
be this plane and 7U the cuspidal tangent. 
In every plane Fe) through PQ (fig. 1), not containing TU, 2 
branches arrive at A from below on // or JV and also 2 from 
1) Somewhat analogous reasoning occurs in comm. 1. to which we do not 
always refer, in order to avoid confusion. 
