odds and ends - proof of lemma 1

lemma 1
Suppose that p,q,r are not on C(a,b,c). Then C(a,b,r),C(b,c,p),C(c,a,q) are concurrent
if and only if H(a,b,c,p,q,r) is real. When they concur, the common point is z≠a,b,c.

Suppose first that the i-lines are pairwise tangent. The contact points must be the obvious
intersections a,b,c, and these are the only intersections. We may apply a mobius
transformation sending b to ∞, and a,c,p,q,r to a',c',p',q',r'. Then the i-lines transform to
parallel lines a'r', c'p' and the circle C(a',c',q') touching the lines at a',c'. Since H is invariant
H(a,b,c,p,q,r) = (a',∞,r',q')(∞'c',p',q')) = (a'-r')(c'-q')/(a'-q')(c'-p')
As a'r',c'p' are parallel lines (a'-r')/(c'-p') is real.
Now, since the lines are parallel, a'c' is a diameter of the circle, <a'q'c' is a right angle
and (c'-q')/a'-q') is purely imaginary. Hence H(a,b,c,p,q,r) is purely imaginary.

It follows that we can concentrate on cases where at least one pair meet twice.
We may as well suppose that C(a,b,r),C(b,c,p) meet twice.
Let z be the intersection of C(a,b,r),C(b,c,p) other than b.
If z = a, then a is on C(b,c,p), so p is on C(a,b,c), a contradiction. Thus z≠a. Similarly z≠c.
Then, as z≠b, (a,b,r,z),(b,c,p,z) are real and non-zero.

The i-lines do not concur at b, or q is on C(a,b,c).
Thus, the i-lines are concurrent if and only if z lies on C(c,a,q), i.e. if and only if
(a,b,r,z)(b,c,p,z)(c,a,q,z) is real. But (a,b,r,z)(b,c,p,z)(c,a,q,z) =H(a,b,c,p,q,r).
The main result follows. The last remark follows from the above proof.

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