for cyclicity, note that $\angle BMP = \angle FEP = \angle PFA$. for second part, note that $PFQD$ is harmonic, and also $EMN \sim FQP$. It is simple from here. $\dfrac{EM}{EN} = \dfrac{FQ}{FP} = \dfrac{BF}{BP}$ since $BQF \sim BFP$

hurdler wrote: for cyclicity, note that $\angle BMP = \angle FEP = \angle PFA$. for second part, note that $PFQD$ is harmonic, and also $EMN \sim FQP$. It is simple from here. $\dfrac{EM}{EN} = \dfrac{FQ}{FP} = \dfrac{BF}{BP}$ since $BQF \sim BFP$ How does $PFQD$ being harmonic is relevant?? Do you need it for something??

Dear Mathlinkers, 1. EF is parallel to BCM 2. According to a converse of Reim's theorem, P, F, B and M are concyclic. Sincerely Jean-Louis

eze100 wrote: hurdler wrote: for cyclicity, note that $\angle BMP = \angle FEP = \angle PFA$. for second part, note that $PFQD$ is harmonic, and also $EMN \sim FQP$. It is simple from here. $\dfrac{EM}{EN} = \dfrac{FQ}{FP} = \dfrac{BF}{BP}$ since $BQF \sim BFP$ How does $PFQD$ being harmonic is relevant?? Do you need it for something?? oh sorry you are right; it is irrevelant. my original solution was longer than it needed to be, and i forgot to edit that out.

hurdler wrote: for cyclicity, note that $\angle BMP = \angle FEP = \angle PFA$. for second part, note that $PFQD$ is harmonic, and also $EMN \sim FQP$. It is simple from here. $\dfrac{EM}{EN} = \dfrac{FQ}{FP} = \dfrac{BF}{BP}$ since $BQF \sim BFP$ hey you mentioned a harmonic range.... please crarify what you mean the points $ P $ , $ F $ , $ Q $ and $ D $ are not collinear..... so perhaps you mean a harmonic pencil...... please specify and use standard notation..... Remark: you don't need that for the proof......

Let $AB$ and $PM$ intersect at $X$. $\angle XFP= \angle PEF= \angle PMB= \angle XMB$ which shows that $BFPM$ is concyclic. We have $\sin \angle BFP= \sin \angle XFP= \sin \angle XMB= \sin \angle EMN$, so $\sin \angle BFP= \sin \angle EMN$ We also have $\sin \angle FPB= \sin \angle FPQ= \sin \angle FEQ= \sin \angle FEN= \sin \angle ENM$. Therefore, $\frac{\sin \angle ENM}{\sin \angle EMN}=\frac{\sin \angle FPB}{\sin \angle BFP}$, so by the law of sines, we have $\frac{EM}{EN}=\frac{BF}{BP}$ as desired. Cheers, Rofler

Dear Mathlinkers, see : http://www.artofproblemsolving.com/Forum/viewtopic.php?t=237263 Sincerely Jean-Louis

Here is my solution : i) Let's show that the points are concylic. It's only angle chasing, thus : . ii) Now we are left to prove that : which is equivalent after applying law of sinus in the two triangles and we have to prove that : . But since : , so it suffices to prove that : , which is true because : . This finish our proof. Q.E.D

My solution Angle chase: ∠AFP=∠PEF=∠PMB (EF//MN) so PFBM is cyclic. Similarity: ∠BNQ=∠FEQ(EF//MN)=∠QFB , so BQFN is cyclic, thus ∠FNE=∠FNQ=∠FBQ=∠FBP, $with \angle FEN=\angle FPB , \triangle FNE \sim \triangle FBP$, so BF/BP=NF/NE M,N are symmetric wrt. AD because ∠FEN=∠EFM (from PFBM is cyclic and EF//MN), so NF=EM , hence BF/BP=EM/EN, as expected

The problem is obvious,and easy to solve with sine theorem.

Part1 : Proving PFBM is cyclic. Note that ∠MBF = ∠EDF = 180 - ∠FPE = 180 - ∠FPM so PFBM is cyclic. Part2 : EM/EN = BF/BP. Note that EM/EN = Sin ∠ENM/Sin ∠EMN = Sin ∠FEQ/Sin ∠PFA = Sin ∠FPB/Sin ∠PFB = BF/BP. we're Done.