odds and ends - proof of wallace's theorem

wallace's theorem
Suppose that z is not on C(a,b,c) and that m≠z. Let p,q,r be the feet of the z-perpendiculars
from m to C(b,c,z),C(c,a,z),C(a,b,z) respectively. Then zεC(p,q,r) if and only if mεC(a,b,c).

Suppose first that m does not lie on C(a,b,z),C(b,c,z) or C(c,a,z), so m≠p,q,r.
By definition, rεC(a,b,z) is such that C(r,m,z) is orthogonal to C(a,b,z).
By lemma 2, this is equivalent to (a,m,z,r) being purely imaginary.
Likewise, (a,m,z,q) is purely imaginary. Thus, (m,a,r,q) = (a,m,z,r)/(a,m,z,q) is real.
Similarly, (m,b,p,r) and (m,c,q,p) are real, so C(a,r,q),C(b,r,p),C(c,p,q) meet in m.
If bεC(p,q,r), then qεC(b,r,p) so C(b,r,p),C(c,p,q),C(a,r,q) meet in q, and therefore
in no other point, so m = q, a contradiction. Thus a,b,c are not on C(p,q,r).
Now miquel's theorem gives the result in each direction.

Now suppose that m is on, say, C(a,b,z). Then r = m.
If m = a, then a = m = r =q. Then p,q(=r) and z must lie on an i-line.
The case m = b is similar, so we may now assume that m≠a,b. Then, since C(a,b,c)
and C(a,b,z) meet only in a,b, m is not on C(a,b,c).
If p is on C(a,b,c), then a is on C(b,c,p)=C(b,c,z), so z is on C(a,b,c), a contradiction.
Similarly, q,r are not on C(a,b,c). Then if zεC(p,q,r), miquel's theorem shows that
mεC(a,b,c), a contradiction.

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