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$ABCD$ একটি বর্গ। $E$ এবং $F$ বিন্দু যথাক্রমে $AB$ এবং $BC$ এর উপর দুইটি বিন্দু যেন $BE = BF$। $B$ বিন্দু হতে $CE$ এর উপর অঙ্কিত লম্ব $CE$ এবং $AD$ কে যথাক্রমে $G$ ও $H$ বিন্দুতে ছেদ করে। $FH$ ও $CE$  রেখা $P$ বিন্দুতে এবং $GF$ and $CD$ রেখা $Q$ বিন্দুতে ছেদ করে। প্রমাণ কর যে, $DP$ ও $BQ$ রেখা পরস্পর লম্ব।


Proof Based Problems  


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Solution

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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.

This is a proof based problem added for learning purposes and does not accept submissions.

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