• It is recommended that you read EGMO (Euclidean Geometry for Mathematical Olympiads) Chapter 1 and 2.
• The most common solution for this problem is the coordinate bash one, but you won't be able to do that if you have 45 minutes left on the clock.

Firstly, we have $\angle EAH = \angle EGH = 90$, hence $AEGH$ is cyclic. We now have, $\angle BHA =\angle GEB = \angle CEB $. Hence $\triangle BAH \cong \triangle BEC$ (ASA Congruency). 

So $AH = BF$ 

$\therefore AHFB$ is a rectangle.

Now, we have $\angle HGC = \angle HFC = \angle HDC = 90$.  So, $HGFCD$ pentagon is cyclic. Also, $\angle BGP = \angle BFP = 90$. So $BGPF$ is also cyclic. 

Now, notice the fact that the radical axises of circles $(HGFCD),(BCD),(BGPF)$ have to be concurrent. Two of these radical axises are $CD$ and $GF$ which are concurrent at $Q$. So the third radical axis or the radical axis of $(BCD),(BGPF)$ must pass through $Q$. Let $R = DP \cap (BGF)$. 

Now we have, 

\[\angle BRD = \angle BRP = \angle BFP = 90\ = \angle BCD \]

This implies that, quad $BRCQ$ is also a cyclic quad . So, $BR$ must be the radical axis. So $B,R,Q$ are collinear. Hence we are done.