Let $(O)$ be the circumcircle of $\triangle ABC.$ $F$ is obviously the midpoint of arc $BAC,$ thus line $FM$ cuts $(O)$ again at the midpoint $D$ of arc $BC,$ i.e. $AD$ bisects $\angle BAC$ $\Longrightarrow$ $AD \parallel MK.$ Therefore $\frac{AK}{FA}=\frac{DM}{FD},$ but power of $M$ WRT $(O)$ is given by $\frac{BC^2}{4}=DM \cdot FM$ $\Longrightarrow$ $\frac{AK}{FA}=\frac{BC^2}{4FM \cdot FD}.$ Since $FA \cdot FP=FM \cdot FD$ $\Longrightarrow$ $BC^2=4PF \cdot AK.$

hello to luis ; can you explain why u say in your first sentense it is obvious that f is the midpoint of arc BAC explain me please enigma

Quote: In acute triangle $ABC$, $c > b$ . Let $M$ be the midpoint of side $[BC]$ . The exterior angle bisector of $\widehat{BAC}$ meet ray $(BC$ at $P$ . The points $K$ and $F$ lie on $PA$ such that $MF \perp BC$ and $MK \perp PA$ . Prove that $4\cdot PF \cdot AK=BC^2$ . Proof 1 (similar with luisgeometria's). Denote the circumcircle $w$ of $\triangle ABC$ . Observe that $F\in w$ . Denote the diameter $[FD]$ of $w$ . Observe that $MK\parallel DA$ and $\widehat{MPA}\equiv\widehat{FMK}\equiv\widehat{MDA}$ $\implies$ $\widehat{MPA}\equiv\widehat{MDA}$ , i.e. $AMDP$ is cyclically. Thus, $\left\|\begin{array}{ccc} MK\parallel DA & \iff & \frac {FA}{KA}=\frac {FD}{MD}\\\\ AMDP\ -\ \mathrm{cyclic} & \iff & FM\cdot FD=FA\cdot FP\\\\ FBDC\ -\ \mathrm{cyclic} & \iff & BC^2=4\cdot MD\cdot FM\end{array}\right\|\ \bigodot$ $\implies\ 4\cdot PF \cdot AK=BC^2$ . Proof 2 I"ll use same notations from upper proof. Denote $L\in AD$ for which $ML\perp AD$ . Observe that $AK=ML$ and $\triangle LMD\sim\triangle MFP$ . Therefore, $\frac {LM}{MF}=\frac {MD}{FP}$ $\iff$ $AK\cdot PF=MD\cdot MF=MB^2$ $\iff$ $4\cdot PF \cdot AK=BC^2$ .

Lemma: Suppose A,C are harmonic conjugates w.r.t B,D. Let M be the midpoint of AC. Then $AM^2=BM*DM$ Proof: Since the guys are harmonic, there exists a point P s.t. $APC=90$, and $PC$ bisects $BPD$. Let Q be on PM s.t. BQ||CP. Since $\Delta PMC$ is iscoselse, quadrilateral $PCBQ$ is also iscoselese. Thus, $DPC=BPC=ACP$, so DP||QC. Now by similar triangles, $MP/MQ=MC/MB=MD/MC$, as desired. [] Let the A internal bisector meet BC at D, then $AD||KM$ and BDCP is harmonic, so by lemma 1, $PM*DM=MC^2$ $4PM*DM=BC^2$ (1) Additionally, by similar triangles, $MD/KA=PM/PK=PF/PM$ (2) Combining (1) and (2) we get $4KA*PF=4PM*MD=BC^2$

Here is my solution: Let $\theta=\angle APB=\frac{1}{2}(C-B)$ It is well-known that $\sin\theta=\frac{c-b}{a}\cos{\frac{A}{2}}$ and $\cos\theta=\frac{c+b}{a}\sin{\frac{A}{2}}$ Sine Rule on $\triangle ACP$ gives: $|CP|=\frac{ab}{c-b}$ Sine Rule on $\triangle ABP$ gives: $|BP|=\frac{ca}{c-b}$ So $|MP|=\frac{|CP|+|BP|}{2}=\frac{a(b+c)}{2(c-b)}$ $|PF|=\frac{|MP|}{\cos\theta}=\frac{a^2}{2(c-b)\sin\frac{A}{2}}$ $|PK|=|MP|\cos\theta=\frac{(b+c)^2\sin\frac{A}{2}}{2(c-b)}$ Sine Rule on $\triangle ACP$ gives: $|PA|=\frac{|CP|\sinh}{\cos\frac{A}{2}}=\frac{4bc\sin\frac{A}{2}}{2(c-b)}$ So $|AK|=|PK|-|PA|=\frac{(c-b)\sin\frac{A}{2}}{2}$ Hence $4\cdot|PF|\cdot|AK|=a^2=|BC|^2$ as required.

enigmation wrote: to luis ; can you explain why u say in your first sentense it is obvious that f is the midpoint of arc BAC explain me please There may be an obvious way that I'm missing. But if you let $O$ be the circumcenter of $\triangle{ABC}$ (which is on line $FM$), then it's easy to get that $\angle{AOM}=2\angle{FAO}$, whence \[\angle{AFO}=\angle{AFM}-\angle{FAO}=\angle{FAO}.\] I have a slightly different solution than the ones above that uses that $ACBF$ is cyclic, and that $PM$ is tangent to the circumcircle of $\triangle{KFM}$. Note that what we need to show is equivalent to \begin{align*} BC^2=4PF\cdot AK=4PF(PK-PA)&=4PF\cdot PK-4PF\cdot PA\\ &=4PM^2-4PF\cdot PA=(PB+PC)^2-4PF\cdot PA, \end{align*}or \[4PF\cdot PA=(PB+PC-BC)(PB+PC+BC)=(2PC)(2PB)=4PC\cdot PB,\]which follows from the power of point $P$ with respect to the circle through $ACBF$.

How about this? $PA\cdot PF=PC\cdot PB=PM^2-MC^2$ and $PK\cdot PF=PM^2$. Subtract and get the desired result.

My Solution Draw the interior angle bisector $AM'$ so we will get that $ AM' \parallel KM $ This implies $ \frac {PK}{AK}=\frac{PM}{MM'} $ so $ \frac {MM'}{AK}=\frac{PM}{PK}=\frac{PF}{PM} $ This implies $PM.MM'=PF.AK$ $PM.MM'= (\frac{a}{2}+\frac{ab}{c-b})(\frac{a}{2}-\frac{ab}{b+c}) =a^2/4$ hence $BC^2=4PF.AK$

Note that BC^2 = (2MC)^2 so we can instead prove MC^2 = PF.AK . we have PM^2 = PA.PF so we can show PM^2 - MC^2 = PA.PF or PC.PB = PA.PF or FACB is cyclic. Let S be reflection of C across FA clearly S lies on DA. we have ∠ACF = ∠ASF and CF = SF = BF so ∠ACF = ∠ABF so FACB is cyclic. we're Done.