Pole. Polar. Theorems of Brianchon and Brokard

\(A\in b\) if and only if \(\angle AB^*O=90^{\circ}\). Analogously \(B\in a\) if and only if \(\angle BA^*O=90^{\circ}\), and it remains to notice that according to the basic properties of inversion we have \(\angle AB^*O=\angle BA^*O\).

Let \(C_1\) and \(D_1\) be the intersection points of \(OA\) with \(k\). Since the inversion preserves the cross-ratio and \(\mathcal R(C_1,D_1;A,A^*)=\mathcal R(C_1,D_1;A^*,A)\) we have \[\mathcal H(C_1,D_1;A,A^*).\quad\quad\quad\quad\quad (7)\] Let \(p\) be the line that contains \(A\) and intersects \(k\) at \(C\) and \(D\). Let \(E=CC_1\cap DD_1\), \(F=CD_1\cap DC_1\). Since \(C_1D_1\) is the diameter of \(k\) we have \(C_1F\bot D_1E\) and \(D_1F\bot C_1E\), hence \(F\) is the orthocenter of the triangle \(C_1D_1E\). Let \(B=EF\cap CD\) and \(\bar{A}^*=EF\cap C_1D_1\). Since \[C_1D_1A\bar{A}^* \frac{E}{\overline\wedge} CDAB \frac{F}{\overline\wedge} D_1C_1A\bar{A}^*\] have \(\mathcal H(C_1,D_1;A,\bar{A}^*)\) and \(\mathcal H(C,D;A,B)\). (7) now implies two facts:

\(1^{\circ}\) From \(\mathcal H(C_1,D_1;A,\bar{A}^*)\) and \(\mathcal H(C_1,D_1;A,A^*)\) we get \(A^*=\bar{A}^*\), hence \(A^*\in EF\). However, since \(EF\bot C_1D_1\), the line \(EF=a\) is the polar of \(A\).

\(2^{\circ}\) For the point \(B\) which belongs to the polar of \(A\) we have \(\mathcal H(C,D;A,B)\). This completes the proof.

We will use the convention in which the points will be denoted by capital latin letters, and their respective polars with the corresponding lowercase letters.

Denote by \(M_i\), \(i=1,2,\dots,6\), the points of tangency of \(A_iA_{i+1}\) with \(k\). Since \(m_i=A_iA_{i+1}\), we have \(M_i\in a_i\), \(M_i\in a_{i+1}\), hence \(a_i=M_{i-1}M_i\).

Let \(b_j=A_jA_{j+3}\), \(j=1,2,3\). Then \(B_j=a_j\cap a_{j+3}=M_{j-1}M_j\cap M_{j+3}M_{j+4}\). We have to prove that there exists a point \(P\) such that \(P\in b_1,b_2,b_3\), or analogously, that there is a line \(p\) such that \(B_1,B_2,B_3\in p\). In other words we have to prove that the points \(B_1\), \(B_2\), \(B_3\) are collinear. However this immediately follows from the Pascal\(\prime\)s theorem applied to \(M_1M_3M_5M_4M_6M_2\).

We will prove that \(EG\) is a polar of \(F\). Let \(X=EG\cap BC\) and \(Y=EG\cap AD\). Then we also have \[ADYF\frac{E}{\overline\wedge} BCXF\frac{G}{\overline\wedge} DAYF,\] which implies the relations \(\mathcal H(A,D;Y,F)\) and \(\mathcal H(B,C;X,F)\). According to the properties of polar we have that the points \(X\) and \(Y\) lie on a polar of the point \(F\), hence \(EG\) is a polar of the point \(F\). Since \(EG\) is a polar of \(F\), we have \(EG\bot OF\). Analogously we have \(FG\bot OE\), thus \(O\) is the orthocenter of \(\triangle EFG\).