# Pole. Polar. Theorems of Brianchon and Brokard

** Definition **

Given a circle \( k(O,r) \), let \( A^* \)
be the image of the point \( A\neq O \)
under the inversion with respect to \( k \).
The line \( a \) passing through
\( A^* \) and perpendicular to \( OA \)
is called the

*polar* of \( A \)
with respect to \( k \).
Conversely \( A \) is called the

*pole* of \( a \)
with respect to \( k \).

** Theorem 6 **

Given a circle \( k(O,r) \), let
and \( a \) and \( b \) be the polars of
\( A \) and \( B \) with respect to \( k \).
The \( A\in b \) if and only if \( B\in a \).

\( A\in b \) 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
\( \angle AB^*O=\angle BA^*O \).

** Definition **

Points \( A \) and \( B \) are called

*conjugated* with respect
to the circle
\( k \) if one of them lies on a polar of the other.

** Theorem 7 **

If the line determined by two conjugated points \( A \) and
\( B \) intersects \( k(O,r) \) at \( C \) and \( D \),
then \( \mathcal H(A,B;C,D) \). Conversely if \( \mathcal H(A,B;C,D) \),
where \( C,D\in k \) then \( A \) and \( B \) are conjugated
with respect to \( k \).

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.

** Theorem 8 (Brianchon) **

Assume that the hexagon
\( A_1A_2A_3A_4A_5A_6 \) is circumscribed about the circle \( k \).
The lines \( A_1A_4 \), \( A_2A_5 \), and \( A_3A_6 \)
intersect at a point.

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 \( M_1M_3M_5M_4M_6M_2 \).

From the previous proof we see that the Brianchon\( \prime \)s theorem
is obtained from the Pascal\( \prime \)s
by replacing all the points with their polars
and all lines by theirs poles.

** Theorem 9 (Brokard) **

The quadrilateral \( ABCD \) is inscribed in the circle
\( k \) with center \( O \). Let \( E=AB\cap CD \),
\( F=AD\cap BC \), \( G=AC\cap BD \).
Then \( O \) is the orthocenter of the triangle
\( EFG \).

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 \( \triangle EFG \).