$EF$ bisects $AD$ $\implies$ $\angle AEF=\angle FED=\angle ABC$, hence, $EF$ is tangent to $\odot (BPED)$, similarly, $EF$ is tangent to $\odot (QDFC)$, now, $$\angle ABQ=180^{\circ}-\angle ACQ=\angle FDQ=\angle EFD+\angle QFC=180^{\circ}-\angle DFQ-\angle AFE=180^{\circ}-\angle EFQ$$hence, $BEFQ$ is cyclic, similarly, $PEFC$ is also cyclic, now notice $EF,PC,BQ$ are radical axes WRT $\odot (PEFC), \odot (BEFQ)$ ; $\odot (PEFC), \odot (ABC)$ and $\odot (ABC) , \odot (BEFQ)$ and hence by Radical Axes Theorem, they all are concurrent

$EF$ bisects $AD$ $\implies$ $\angle AEF=\angle FED=\angle ABC$, hence, $EF$ is tangent to $\odot (BPED)$, similarly, $EF$ is tangent to $\odot (QDFC)$, now, $$\angle EPC=\angle PEK+\angle PCB=\angle EBP+\angle PCB=\angle ACB=\angle AFE$$hence, $BEFQ$ is cyclic, similarly, $PEFC$ is also cyclic, now notice $EF,PC,BQ$ are radical axes WRT $\odot (PEFC), \odot (BEFQ)$ ; $\odot (PEFC), \odot (ABC)$ and $\odot (ABC) , \odot (BEFQ)$ and hence by Radical Axes Theorem, they all are concurrent

Just a little different

dangerousliri wrote: Let $ABC$ be an acute triagnle with its circumcircle $\omega$. Let point $D$ be the foot of triangle $ABC$ from point $A$. Let points $E,F$ be midpoints of sides $AB,AC$, respectively. Let points $P$ and $Q$ be the second intersections of of circle $\omega$ with circumcircle of triangles $BDE$ and $CDF$, respectively. Suppose that $A,P,B,Q$ and $C$ be on a circle in this order. Show that the lines $EF,BQ$ and $CP$ are concurrent. Could anybody show that also the tangent of $\omega$ at point $A$ pass through this intersection of this lines.

dangerousliri wrote: dangerousliri wrote: Let $ABC$ be an acute triagnle with its circumcircle $\omega$. Let point $D$ be the foot of triangle $ABC$ from point $A$. Let points $E,F$ be midpoints of sides $AB,AC$, respectively. Let points $P$ and $Q$ be the second intersections of of circle $\omega$ with circumcircle of triangles $BDE$ and $CDF$, respectively. Suppose that $A,P,B,Q$ and $C$ be on a circle in this order. Show that the lines $EF,BQ$ and $CP$ are concurrent. Could anybody show that also the tangent of $\omega$ at point $A$ pass through this intersection of this lines. Let $Z$ be the intersection point of $EF$ , $PC$ , $BQ$.Now we prove that $(BQFE)$ cyclic from this we have that $ZE$ x $ZF$=$ZB$ x $ZQ$ that is the power of Z whith respect to $(ABC)$ from this we conclude that $AZ$ is tangent to $(ABC)$