Property 1:-

Given a triangle $ABC$ , suppose $S$ is its symmedian point, then the tri-linear polars of all points on the circumcircle of $ABC$ passes through $S$ .

Proof:-

Suppose, $P$ is a point on $\odot ABC$ . Tri-linear polar of $P$ be $A_1B_1C_1$ . $A'B'C'$ be the tangential triangle of $ABC$ and $P_aP_bP_c$ be the cevian triangle of $P$ wrt $ABC$ . Then note that by Pascal's Theorem $P_b,P_c,A'$ are collinear. If $A_2=AA'\cap BC$ . Then note that $(AA_2;SA')=-1$ and we have $(AB;P_cC_1)=-1$ and $(AC;P_bB_1)=-1$ . So $B_1,C_1,S$ are collinear. So done.

Property 2:-

In a triangle $ABC$ , $A_1B_1C_1$ is the tri-linear polar of a point $P$ and $G$ is the centroid of $ABC$ . Then if the tri-linear polars pass through some fixed point, then $P$ moves on a conic passing through $A,B,C,P$ .

Proof:-

Note that this proof directly follows from property 1, if we take that fixed points to the symmedian point of $ABC$ by a projective transformation.

Equivalent Problem:-

If the fixed point is at infinity, i.e. if the tri-linear polars are parallel, then $P$ moves on the hyperbola passing through $A,B,C,P,$ and the centroid of $ABC$ .

Property 3:-

In a triangle $ABC$ , isotomic points of all the points on the circumcircle of $ABC$ lie on the isotomic line of Lemoine axis.

Proof:-

From property 10 of the previous post, we can see that the isotomic points are collinear since the isogonal conjugate of the circumcircle is the line at infinity.

Property 4:-

In a triangle $ABC$ , if a point $P$ moves on a line then the isotomic point of $P$ and also the isogonal conjugate of $P$ moves on two hyperbola passing through $A,B,C$ .

Proof:-

By a projective transformation take the line to infinity. But the isogonal conjugate of $P$ doesn't remain its isogonal conjugate wrt the new triangle. To recover the isogonal property, just note that for any point $X$ on $BC$ , after the projective transformation the ratio $\frac {BX}{XC}$ has been multiplied by some factor. So by an affine transformation if we multiply the lengths of the sides by proper ratio, then the isogonal property will recovered(because affine transformation preserves ratios). So now locus of the isogonal conjugate is the circumcircle of $ABC$ . So it was a hyperbola passing through $A,B,C$ before. The isotomic case is trivial from this becuase the composition of isogonal and isotomic transformation takes a line to a line.

Property 5:-

The isotomic points of the tri-linear poles of two parallel lines are collinear with the centroid of the triangle.

Proof:-

Suppose, $l_1,l_2$ are the tri-linear polars of $P,Q$ . Isotomic points of $P,Q$ be $P',Q'$ . Since $l_1\parallel l_2$ . So from property 2 $A,B,C,P,Q,G$ lie on the same hyperbola where $G$ is the centroid of $ABC$ . So from property 4, $P',Q',G$ are collinear.

Property 6:-

The isogonal conjugates of the tri-linear poles of two parallel lines are collinear with the symmedian point of the triangle.

Proof:-

It directly follows from property 5 if we use the property 10 of previous post.

Property 7:-

In a triangle $ABC$ , $H$ is its orthocenter. $P$ is a point on the plane. The perpendiculars from $H$ to $AP,BP,CP$ intersect $BC,CA,AB$ at $A_1,B_1,C_1$ . Then $A_1,B_1,C_1$ are collinear on a line perpendicular to $HP$ and if $Q$ is the tri-linear pole of $A_1B_1C_1$ , then $HQ$ is perpendicular to the tri-linear polar of $P$ .

Proof:-

If $A_2,B_2,C_2$ are the feet of the perpendicular from $H$ to $AP,BP,CP$ and $A'B'C'$ be the orthic triangle of $ABC$ , then $HA.HA'=HA_2.HA_1$ and similar for others. Note that, $A_2,B_2,C_2$ lie on the circle with diameter $HP$ . So under inversion wrt $H$ with power $HA.HA'$ , $A_2,B_2,C_2$ go to $A_1,B_1,C_1$ . So $A_1,B_1,C_1$ are collinear on a line perpendicular to $HP$ . Suppose, $A_3B_3C_3$ is the tri-linear polar of $P$ wrt $ABC$ and $P_aP_bP_c$ be its cevian triangle. $H$ is the orthocenter of $\Delta AP_aA_1$ . So $HP_a\perp AA_1$ . But $(HB,HC;HP_a,HA_3)=-1$ and $(AC,AB;AQ,AA_1)=-1$ . Since $HB\perp AC, HC\perp AB, HP_a\perp AA_1$ . So $HA_3\perp AQ$ and similar for others. So similarly we get $HQ\perp A_3B_3C_3$ . So done.

Orthogonal Conjugate:-

I will call the points $P,Q$ in Property 7 orthogonal conjugates to each other.

Property 8:-

If a point moves on a line its orthogonal conjugate moves on a hyperbola.

Proof:-

Clearly, if a point moves on a hyperbola, its tri-linear polars are concurrent. But note that if $-k^2$ is the power H wrt the nine-point circle of the triangle, then the feet ofthe perpendiculars from $H$ to these plars go to the orthogonal conjugates of corresponding points under inverting with this power. Since initially the feet of the perpendiculars lie on a circle passing through $H$ , so under inversion they go to a line. So done.

Property 9:-

In a triangle $ABC$ , $H$ is the orthocenter of $ABC$ , $P$ be a point on the Euler line of $ABC$ . Suppose, $A_1B_1C_1$ is the tri-linear polar of its orthogonal conjugate. Then the isotomic line of $A_1B_1C_1$ wrt $ABC$ is perpendicular to $HP'$ where $P'$ is the isogonal conjugate of $P$ wrt $ABC$ .

Proof:-

Suppose, the tri-linear polar of the orthogonal conjugate of $P'$ is $A_2B_2C_2$ . $A_3B_3C_3$ be the isotomic line of $A_1B_1C_1$ wrt $ABC$ . Suppose, the tangents on $\odot BHC,\odot CHA,\odot AHB$ at $H$ intersect $BC,CA,AB$ at $A_4,B_4,C_4$ . Note that, $A_4B_4C_4$ is perpendicular to the Euler line of $ABC$ and also from Property 7 $A_1B_1C_1$ is also perpendicular to the Euler line. So $A_1B_1C_1\perp A_4B_4C_4$ . Now note that $(BC;A_3\infty)=(BC;A_2A_4)=(BC;\infty A_1)$ and similar for others. So from property of previous post we get that $A_2B_2C_2\parallel A_3B_3C_3$ . So done.

Property 10:-

In the previous problem, $HP'$ is also perpendicular to the tri-linear polar of $P$ wrt $ABC$ .

Proof:-

Suppose, $A_4B_4C_4$ is the tri-linear polar of $H$ wrt $ABC$ . Clearly, $A_4B_4C_4$ is perpendicular to the Euler line of $ABC$ . $A_5B_5C_5$ be the tri-linear polar of $P$ wrt $ABC$ . $A_6$ be the foor of the perpendicular from $A$ to $BC$ and $A_7$ be the harmonic conjugate of $A_6$ wrt $BC$ . Now just note that, $(BC,A_1\infty)=(HB,HC;HA_1,BC)=(AC,AB;AP,BC)=$ $-(CB;A_5\infty)$ . Using this we can easily get that, $ (BC;A_7A_5)=(BC;A_1\infty)=(BC;\inftyA_3) $. So uding property 10 of previous post we get that $A_5B_5C_5\parallel A_3B_3C_3$ . So using previous property we get that $HP'\perp A_5B_5C_5$ .

Some consequences:-

1. Note that orthogonal conjugate of the Euler line is the hyperbola passing through $A,B,C,H$ and the centroid( $G$ ) of $ABC$ .

2. Isotomic conjugate of of this hyperbola is the line passing through $G$ and the symmedian point( $S$ ) of $ABC$ .

3. Note that in property 9 and 10, suppose, $Q$ is the orthogonal conjugate of $Q$ and $Q'$ be the isotomic conjugate of $Q$ wrt $ABC$ . Clearly, $Q'$ lies on $GS$ . $HQ$ is perpendicular to $A_5B_5C_5$ . But $A_5B_5C_5\parallel to A_3B_3C_3$ which is the tri-linear polar of $Q'$ . So we can conclude that if $Q$ lies on $GS$ then the line joining $H$ and the isotomic point of $Q$ is perpendicular to the tri-linear polar of $Q$ .

4. Easy to see that the previous property is an "if and only if" property

5. For any point $P$ on the Euler line its orthogonal conjugate $Q$ and its isogonal conjugate $P'$ are collinear with $H$ .

6. Isotomic points of the Fermat points of a triangle lies on $GS$ .

Proof:-

Note that orthocorrespondent of the Fermat points are the Fermat points themselves. But we know that the line joining orthocenter and any point is perpendicular to the isotomic line of its orthocorrespondence. So the line joining orthocenter and Fermat point is perpendicular to the isotomic line of the tri-linear polar of the Fermat point. So from the previous property we get that isotomic points of the Fermat points lie on $GS$ .

This property can also be proved by using property 10 of previous post and that the two isodynamic points lie on the Brocard axis.

Property 11:-

The projective transformation followed by an affine transformation that takes a point $P$ to $Q$ keeping the triangle $ABC$ fixed, takes $Q'$ to $P'$ where $Q',P'$ are the isogonal conjugates of $Q,P$ wrt $ABC$ .

Proof:-

Suppose, under the transformation $Q'$ goes to $K$ . So $(AB,AC;AP,AQ')=(AB,AC;AQ,AK)$ since the transformation preserves cross-ratio. Also $(AB,AC;AP,AQ')=(AC,AB;AP',AQ)$ since reflecting on the angle-bisector of $\angle BAC$ doesn't change the cross-ratio of the lines. And $(AC,AB;AP',AQ)=(AB,AC;AQ,AP')$ . Sp $AP'\equiv AK$ . Similarly $BP'\equiv BK$ . So $K\equiv P'$ . So done.

Property 12:-

Given two points $P,Q$ and their isogonal conjugates wrt $ABC$ be $P',Q'$ , then suppose, $X=PQ'\cap QP'$ and $Y=PQ\cap P'Q'$ , then $Y$ is the isogonal conjjugate of $X$ wrt $ABC$ .

Proof:-

So if the transformation in property 11 takes $P'$ to $Q$ , then it takes $Q'$ to $P$ . So $(P'A,P'B;P'C,P'Q')=(QA,QB;QC,QP)$ $\implies (P'A,P'B;P'C,P'Y)=(QA,QB;QC,QY)$ . So $A,B,C,Q,Y,P'$ lie on the same conic, which is the isogonal conjugate of $PQ'$ wrt $ABC$ . So isogonal conjugate of $Y$ lies on $PQ'$ and similarly it lies on $QP'$ . So $X$ is the isogonal conjugate of $Y$ .

Property 13:-

Given a triangle $ABC$ and its centroid $G$ , let $A'B'C'$ be the anti-cevian triangle of a point $P$ , $Q$ be a moving point. Through $Q$ draw a line parallel to $B'C'$ to intersect $BC$ at $A_1$ . Similarly define $B_1,C_1$ . Prove that $A_1,B_1,C_1$ are collinear iff $Q$ lies on the hyperbola passing through $A,B,C,P$ and the anti-compliment of $P$ .