This is brokards theorem.

simple, it's polar line.

can somebody please post a full proof. Tx

Pretty well-known. A concise proof can be found here(theorem 9).

It is trivial by the concept of poles and polars.we get, polar of $P$ passing through $Q$ as $P=AB \cap CD$.so from La Hire's theorem we get, $P$ is on the polar of $Q$. and $EF$ is the polar of $Q$(tangent).so $P$ is on the line $EF$.Hence $P,E,F$ are collinear.

Dear Mathlinkers, for history and more, you can see http://perso.orange.fr/jl.ayme vol. 13 La réciprocité polaire de Philippe de La Hire p. 21-24. Sincerely Jean-Louis

I don't understand why a direct application of Brocard's theorem is posed.It is well known that $PR$ is the polar of $Q$ where $R=AC \cap BD$.Again $EF$ is the polar of $Q$.The result follows....

Solution with Complex Numbers WLOG it is unit circle. Lemma 1: A quadrilateral $ABCD$ is inscribed in unit circle.If $P=AB\cap CD$ and $Q=AD\cap BC$, then \[p\cdot \overline{q}+q\cdot\overline{p}=2\]Proof: See here Back to the problem We have $QE=QF$ $\to$ $|q-f|^2=|q-e|^2\to$ $$(q-f)(\overline{q}-\overline{f})=(q-e)(\overline{q}-\overline{e})$$We get $ef=\frac{q}{\overline{q}}$. We want to show that \[\frac{e-f}{\overline{e}-\overline{f}}=\frac{e-p}{\overline{e}-\overline{p}}\]$\to$ \[\frac{e+f}{ef}=\overline{p}+\frac{p\overline{q}}{q}\]Since $q=\frac{2ef}{e+f}$ we must prove \[\frac{2}{q}=\overline{p}+\frac{p\overline{q}}{q} \ \to p\cdot \overline{q}+q\cdot\overline{p}=2\]The last one is Lemma 1.

By brokards theorem, the polar of $Q$ pass through $P$. And it's known that, $\overleftrightarrow{EF}$ is the polar of $Q$. So, the polar of $Q$ pass through $P, E, F$. In other word $P,E,F$ are collinear. $\mathbb Q. \exists. \mathbb D$

Attachments:

Is this solution correct? Let $AC \cap BD=R$, then if, $PR \cap BQ=G$, then, $-1 = (B,C;G,Q)$, now note that, $EBFC$ is harmonic quadrilateral, hence if $EF \cap BQ =X$, then, $-1=(B,C;X,Q)$ $\implies$ $\boxed{X=G}$ $\implies $ $P-E-F$

Sorry if I have identical solution with the others but I had to do it. By La Hire's theorem $P$ lies on the polar of $Q$ but $EF \perp OQ'$ so $EF$ is the polar of $Q$. Hence $P, E, F$ are collinear.

An unnecessarily complicated solution!!! Chinese MO 1997 P4 wrote: Consider a cyclic quadrilateral $ABCD$. The extensions of its sides $AB,DC$ meet at the point $P$ and the extensions of its sides $AD,BC$ meet at the point $Q$. Suppose $\quad QE,QF$ are tangents to the circumcircle of quadrilateral $ABCD$ at $E,F$ respectively. Show that $P,E,F$ are collinear. Solution: Let $\odot(ABCD)=\omega$ with center $O$. After performing an inversion about circle with center $Q$ and radius $QF=QE$ and then performing another inversion to the resultant diagram about $\omega$ the problem is equivalent to: Inverted Problem wrote: Consider a cyclic quadrilateral $ABCD$. Let $CD\cap AB=Q$. Let $E,F$ be tangency points of tangents from $Q$ to $\omega$. Prove that $EF,AC,BD$ concur. Notice that this problem is purely projective. Perfoming a projective transformation $\mathcal{P}$ that fixes $\omega$ and sends $AC\cap BD$ to center of $\omega$. Let $X'$ denote the mapped point to $X$ under $\mathcal{P}$. So, $A'B'C'D'$ is rectangle and the tangents at $E',F'$ are parallel (as $A'B'\parallel C'D'$). Hence, $A'C',B'D',E'F'$ concur (at center of $\omega$). $\square$ Hence, inverting back we get the desired result. $\blacksquare$

by Brocard’s Theorem: Let $R=AC \cap BD$($\triangle PQR$ is autopolar) Then: $q \equiv RP$ But $EQ$ and $QF$ are tangents to $\omega$ in $E$ and $F$, respectively Then $q \equiv EF$ $\implies E,R,F,P$ are collinearity $\implies P,E,F$ are collinear.

jred wrote: Consider a cyclic quadrilateral $ABCD$. The extensions of its sides $AB,DC$ meet at the point $P$ and the extensions of its sides $AD,BC$ meet at the point $Q$. Suppose $\quad QE,QF$ are tangents to the circumcircle of quadrilateral $ABCD$ at $E,F$ respectively. Show that $P,E,F$ are collinear. Too easy for China By Brokard $P \in \text{Polar}_Q$ which is $EF$.

Brocards : Let $R=AC \cap BD$($\triangle PQR$ is autopolar) Q is the pole of line EF . Since P also lies on the polar of Q. We are done