Triangle Conjugacy is Mutual

Theorem

Let $\CC$ be a circle.

Let $\triangle PQR$ be a triangle.

Let $\triangle P'Q'R'$ be such that:

$P'$ is the pole of $QR$

$Q'$ is the pole of $PR$

$R'$ is the pole of $PQ$

with respect to $\CC$.

Then:

$P$ is the pole of $Q'R'$

$Q$ is the pole of $P'R'$

$R$ is the pole of $P'Q'$

with respect to $\CC$.

That is, $\triangle PQR$ and $\triangle P'Q'R'$ are conjugate triangles with respect to $\CC$.

Proof

We have that:

the polar of $P'$ is $QR$

the polar of $Q'$ is $PR$

and so both polars pass through $R$.

Therefore:

the polar of $R$ is $P'Q'$.

Similarly:

the polar of $P$ is $Q'R'$

and:

the polar of $Q$ is $P'R'$.

$\blacksquare$

Sources

• 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {III}$. The Circle: $9$. Conjugate triangles