Easy Problem in geo, (cyclic, angle chasing)

[Kazi_Zareer]

$ABCD$ is a cyclic quadrilateral, with diagonals $AC,BD$ perpendicular to each other. Let point $F$ be on side $BC$, the parallel line $EF$ to $AC$ intersect $AB$ at point $E$, line $FG$ parallel to $BD$ intersect $CD$ at $G$. Let the projection of $E$ onto $CD$ be $P$, projection of $F$ onto $DA$ be $Q$, projection of $G$ onto $AB$ be $R$. Prove that $QF$ bisects $\angle PQR$. (Source:china TST 2014 P1)

[nahin munkar]

Here,we can show $AQYR$ & $XQDP$ is cyclic. So,By easy angle chasing, we can show $\angle RQF$ =$\angle FQP$.That's the proof.[here, RG cuts AC at Y & EP cuts BD at X. It's a simple proof.Try to yourself.]. Proved.

[tanmoy]

Apply a projective transformation that sends $ABCD$ to a square $A_{1}B_{1}C_{1}D_{1}$.The rest is very easy

[nahin munkar]

Well done.It's another good approach in projective way.

[Kazi_Zareer]

Here,we can show $AQYR$ & $XQDP$ is cyclic. So,By easy angle chasing, we can show $\angle RQF$ =$\angle FQP$.That's the proof.[here, RG cuts AC at Y & EP cuts BD at X. It's a simple proof.Try to yourself.]. Proved.

I have already solved it and given the main clues of my solution.

[Kazi_Zareer]

Apply a projective transformation that sends $ABCD$ to a square $A_{1}B_{1}C_{1}D_{1}$.The rest is very easy

Awesome sala!

[nahin munkar]

Here,we can show $AQYR$ & $XQDP$ is cyclic. So,By easy angle chasing, we can show $\angle RQF$ =$\angle FQP$.That's the proof.[here, RG cuts AC at Y & EP cuts BD at X. It's a simple proof.Try to yourself.]. Proved.

I have already solved it and given the main clues of my solution.

Yes,dude.I solved from those 2 clues-(cyclic,angle-chasing).