hmm...seems to be evident as well. since $ QFPD$ is a harmonic quadrilateral,it follows that $ MD = DN$.Therefore, $ P,F,N$ are collinear,it follows that $ \angle FBP = \angle PMF$ due to symmetry and a really simple angle chasing. Recalling the property of the harmonic quadrilateral,$ QFBD$, we obtain that $ \boxed{BD = BN}$. Simple angle chasing yields that $ \triangle BPN\sim\triangle EMN$,so combining it with the boxed quality we are finishing the proof.

There is clearly a typo with the ratio equality in Part 2 of the question, but I can't figure what. Part 1 is correct though. Connect $ DE$ and $ DR$ and $ PF$. It remains to show that $ \angle EDF = \angle MBF$. But $ \Delta DBF$ and $ \Delta CDE$ are congruent and isoceles, so $ \angle EDF = 180 - \angle FDB - \angle EDC = 180 - 2 \angle FDB = \angle DBF = \angle MBF$.

Edit: Sorry, bad error by GSP. Click to reveal hidden text

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according to what KenHungKK wrote:$ P$ is a point on arc $ EK$ opposite $ D$, it seems to me that he meant to write that $ P$ and $ D$ are diametrical opposite points,otherwise,the part would not be correct...

It still fails even if $ P$ and $ D$ are diametrically opposite. Refer to the attached .png. Edit: Attached .png was an erroneous GSP sketch as well.

As the question asks us to prove $ \frac{EM}{EN} = \frac{BD}{BP}$, I believe that you'd better compute $ \frac{EM}{EN} \times \frac{BP}{BD}$ but not $ \frac{EM}{EN} \times \frac{BD}{BP}$ as in your diagram

@brian: In my diagram I had computed $ \frac{EM}{EN} - \frac{BD}{BP}$, the - is a tad bit small. Sorry for the confusion, apparently my copy of GSP is messed up. In particular my computations indicated that $ QPFD$ was not harmonic unless $ \Delta ABC$ was equilateral.

as a conclusion: so at long last my solution isn't completely correct, unless points $ P,D$ are diametrically opposite.let's just wait till KenHungKK will clarify the statement. on the other hand if they are not opposite, the statement of the second part does not hold as well, so I'm inclined to suspect that my solution is correct and "satisfies" to the true statement...

Actually a very similar problem appeared long time ago http://www.mathlinks.ro/viewtopic.php?p=1251781&search_id=1803384435#1251781

I have an elementary solution. Part 1 (For completeness): Connect $ DE$ and $ DR$ and $ PF$. It remains to show that $ \angle EDF = \angle MBF$. But $ \Delta DBF$ and $ \Delta CDE$ are congruent and isoceles, so $ \angle EDF = 180 - \angle FDB - \angle EDC = 180 - 2 \angle FDB = \angle DBF = \angle MBF$. Part 2: Let us note 2 results: (Bear in mind $ PFBM$ is concyclic.) 1. $ \Delta EMN \backsim \Delta FQP$. Firstly $ \angle MEN = \angle QFP$. Next, $ \angle MEC = \angle AEP = \angle EQP = \angle EDP$, by Alternate Segment Theorem. Then $ \angle EMN = \angle ECD - \angle MEC = \angle EDF - \angle MEC = \angle EDF - \angle EDP = \angle PDF = \angle PQF$. 2. $ \Delta FQB \backsim \Delta BFP$. This is because $ \angle FBP = \angle FBQ$ (trivially), and $ \angle BFQ = \angle BPF$ by Alternate Segment Theorem. By 1, $ \frac {EM}{EN} = \frac {QF}{PF}$. By 2, $ \frac {QF}{PF} = \frac {BF}{BP}$. But $ \frac {BF}{BP} = \frac {BD}{BP}$, we are done. An (accurate) diagram is attached for reference. For the record: the question KenHungKK posted is correct. We do not require $ P$ and $ D$ to be diametrically opposite.

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for the firs part even faster: since ABC is isoceles $ EF||BC$ and then $ \angle PFA=\angle PEF=\angle PMB$ and we are done

NEW WAY TO SOLVE IT

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