A challanging problem

[sourav das]

Problem:
Let $ABC$ be an acute angled triangle with pedal triangle $DEF$. Let $P$ be the intersection of $AD$ and $EF$ and $Q$,$R$ the intersections of $AB$,$AC$ respectively with the perpendicular bisector of $PD$.
Prove that $ARDQ$ is a cyclic quadrilateral.

Please be honest and reply with solution that how much time you spend to solve this problem.[Or at least try until you surrender]

[*Mahi*]

Honestly , you call this one challenging!
Solution:

solve.png (17.84KiB)Viewed 2520 times

I changed the name of points as per the figure.Let $K$ be the orthocentre of $\triangle ABC$.
$AF \perp BC, JK \perp AF $
So, $JK || BC$
So,$ \triangle ABC $ & $ \triangle AJK $ is homothtical ,which takes the orthocentre $K$ of $ABC$ to point $G$ (consider $\triangle ACE$).
Now it is more than easy to see that $\angle JFK=\angle JGK =\angle BKC =180 -A$
So $AJFK$ is cyclic.
It took 7 minutes ,if you want to know.

[sourav das]

Really a nice solution. But i can't understand how to take orthocenter $K$ of $ABC$ to point $G$(Even if $\Delta ABC$ and $\Delta AJK$ are homothetic) . And what about $\Delta ACE$ ? Isn't it just a line according to your figure? Really i am not so talent as you. So it will be much better if you describe it in details.