**Theorem:** *Any point conic also arises as a line conic*.

Proof: We select four points on our conic: *A*, *B*,
*C*, and *D*, and take tangents to all of them: *T(A)*, *T(B)*, *T(C)*, and
*T(D)*. Then we define:

*B1=T(A).T(B)*; *C1=T(A).T(C)*; *D1=T(A).T(D)*;

*A2=T(B).T(A)*; *C2=T(B).T(C)*; *D2=T(B).T(D)*.

Now *A*, *B1*, *C1*, and *D1* all lie on
*T(A)*, as *A2*, *B*, *C2*, and *D2* all lie
on *T(B)*. Note also that *A2* is the same point as *B1*,
and lies on both *T(A)* and *T(B)*.

We need to prove that *(A,B1;C1,D1) = (A2,B;C2,D2)*. When we have
this, we can fix a projectivity from *T(A1)* to *T(B2)* by
*A->A2*, *B1->B*, *C1->C2*. Then given any
other point *D* on the conic, we can see that its tangent was a
constituent of the line conic because this projectivity took *D1*
to *D2*--i.e. *(A,B1;C1,D1) = (A2,B;C2,D2)*.

Proof of equality of cross ratios: Set up the diagonal triangle
for *A*, *B*, *C*, *D*; let *P=AD.BC*,
*Q=AC.BD*, and *R=AB.CD*. Then the above theorem tells us
that *A2=B1* lies on *PQ*; *C1* and *D2* lie on
*PR*; *C2* and *D1* lie on *QR*. Now we set up a
projectivity through *R*, and find that *(A,B1;C1,D1) =
(B,A2;D2,C2)*. Since we also know that *(B,A2;D2,C2) =
(A2,B;C2,D2)*, we indeed have the desired result: *(A,B1;C1,D1) =
(A2,B;C2,D2)*.

The converse of this theorem is its dual. We could prove it using the duals of all the results in this proof; in particular, we would begin with a special case of Brianchon's theorem in which a quadrilateral is circumscribed to a conic.

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Created: Nov 30 1995 ---
Last modified: Thu Nov 30 15:25:18 1995