Use radical axis to prove three lines are concurrent (problem in image) Please help me tackle this problem. All I see is the trivial fact that EPFG is cyclic. I know I have to show that these lines are radical axis for three circles but I have no idea which ones. Any hints would be appreciated
 A: I don't guarantee that this is the most optimal solution, but here is one that quickly came to mind and justifies the fact you want to prove. 
Let $k$ be the circle circumscribed around rectangle $ABCD$. 
Lemma 1. Let $k_1$ be the circumcircle of triangle $PFQ$ and let $R$ be the second intersection point of $k_1$ and $k$, different from the other intersection point $P$. Then the three lines $BC, \, FQ$ and $PR$ intersect at a common point $H$ and $\angle \, PRQ = 90^{\circ}$.

Proof: Since by the construction of point $R$, quad $PQRF$ is inscribed in the circle $k_1$ so $$\angle \, PRQ = \angle \, PFQ = 90^{\circ}$$ Furthermore, $$\angle \, BFQ =  90^{\circ} = \angle \, BCD = \angle \, BCQ$$ so quad $BFCQ$ is inscribed in a circle $k_2$. Thus, the three lines $BC, \, FQ$ and $PR$ are the radical axes of the corresponding pairs of circles formed by $k_1, \, k_2$ and $k$ and by the three radical axes theorem $BC, \, FQ$ and $PR$ intersect in a common point, denoted by $H$. 
Lemma 2. The three lines $AD, \, EQ$ and $PR$ intersect at a common point, which is $G = EQ \cap AD$.

Proof: Since by assumption $EQ$ is perpendicular to $AP$ and $E = EQ \cap AP$, 
$$\angle \, PEQ = 90^{\circ} = \angle \, PFQ$$ so point $E$ lies on circle $k_1$. Furthermore, $\angle \, AEQ = 90^{\circ} = \angle \, ADQ$ so quad $AEQD$ is inscribed in a circle $k_3$. Consequently, the three lines $AD, \, EQ$ and $PR$ are the radical axes of the corresponding pairs of circles formed by $k_1, \, k_3$ and $k$ and by the three radical axes theorem $AD, \, EQ$ and $PR$ intersect in a common point, which is denoted by $G$.
Corollary. The three lines $BC, \, PG$ and $FQ$ are concurrent.
Proof: By Lemma 2, line $PG$ passes through the point $R \in k_1 \cap k$, so in fact line $PG$ is the same as line $PR$. However, by Lemma 1, line $PR$ passes through point $H = BC \cap FQ$. Consequently, $PG$ passes through $H$, which proves the corollary and solves the problem.   
