$$\frac{AN}{NC} = \frac{AD}{DC} = \frac{AD}{DB} = \frac{AM}{MB} \Longrightarrow MN \parallel BC$$$$\frac{PM}{AP} = \frac{MO}{ON} \cdot \frac{NF}{FA} = \frac{NO}{OM} \cdot \frac{ME}{EA} = \frac{NR}{AR} \Longrightarrow MN \parallel PR$$. $$\frac{AP}{MP} \cdot \frac{MO}{ON} \cdot \frac{NF}{FA} =1 \Longrightarrow \frac{AP}{MP} = \frac{FA}{NF}$$$$\frac{AM}{MP} = \frac{AP}{MP} -1 = \frac{FA}{NF} -1 = \frac{NC}{NF}$$$$1+ \frac{NF}{NC} = 1+ \frac{MP}{AM} \Longrightarrow \frac{FC}{NC} = \frac{AP}{AM} ( \star )$$$$\frac{AP}{AB} = \frac{PR}{BC} ( \star \star )$$$$\frac{BM}{BD} = \frac{AM}{AD} ( \star \star \star )$$Multiplying $( \star )$, $( \star \star )$ and $( \star \star \star )$ we get $\frac{AC}{AB} \cdot \frac{BM}{CN} =1 = \frac{PR}{AD}$ We conclude $PR=AD$

Dear Mathlinkers, an original proof of this problem 6 can be seen on http://jl.ayme.pagesperso-orange.fr/Docs/LIII%20OME%20Problema%206.pdf and http://personal.us.es/rbarroso/trianguloscabri/ Sincerely Jean-Louis

Claim

Let $X$ be a point on $MN$ such that $(M,N;O,X)=-1$.Define $O',X_3,X_2$ as the midpoint of $MN$,the intersection of $RX$,$AB$ and $PX$,$AC$.$P(M,N;O,X)=(A,N;F,X_2)=-1$ and similarly $R(M,N;O,X)=(A,M;E,X_3)=-1$. So $\frac{FA}{FN}=\frac{AX_2}{NX_2}$ and also $\frac{EA}{EM}=\frac{AX_3}{MX_3}$.From Claim and Thales we get $MN\parallel X_2 X_3$.Now let $K$ be the infinity point on $MN$. Clear that $X_3 (M,N;O',K)=(A,N;F',X_2)=(A,N;F,X_2)=-1$ $\implies$ $F'\equiv F$ (here $F'=X_3 O'\cap AC$) which is a contradiction as $X_3$ and $F$ inside of $AM$ and $AN$ then The intersection of $X_3 F$ and $MN$ is outside of $MN$. So $X$ is in infinity and $O$ is midpoint of $MN$ now do an easy side-chasing to get $PR=AD$ this part is easy.

Dear MLs, It is a PWW problem. See attachment. M.T.

Attachments:
ML1432843.doc (29kb)

This can be done easily with barycentric coordinates as well if you're lazy. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.935641238791054, xmax = 26.38171993780454, ymin = -6.874264084978881, ymax = 9.853827672436813; /* image dimensions */pen zzttqq = rgb(0.6,0.2,0); draw((-2.32,6.98)--(-4.98,-0.8)--(5.84,-0.72)--cycle, linewidth(2) + zzttqq); /* draw figures */draw((-2.32,6.98)--(-4.98,-0.8), linewidth(2) + zzttqq); draw((-4.98,-0.8)--(5.84,-0.72), linewidth(2) + zzttqq); draw((5.84,-0.72)--(-2.32,6.98), linewidth(2) + zzttqq); draw((-2.32,6.98)--(0.43,-0.76), linewidth(2)); draw((-3.9237156783181897,2.289433091234769)--(0.43,-0.76), linewidth(2)); draw((0.43,-0.76)--(2.5996691485249728,2.3376651417105037), linewidth(2)); draw((-3.9237156783181897,2.289433091234769)--(2.5996691485249728,2.3376651417105037), linewidth(2)); draw((-4.3392876182874085,1.0739632818511171)--(3.8745063779042277,1.1346937365364522), linewidth(2)); draw((1.76,3.13)--(-4.3392876182874085,1.0739632818511171), linewidth(2)); draw((-3.65,3.09)--(3.8745063779042277,1.1346937365364522), linewidth(2)); /* dots and labels */dot((-2.32,6.98),dotstyle); label("$A$", (-2.2503873233322977,7.146792076838422), NE * labelscalefactor); dot((-4.98,-0.8),dotstyle); label("$B$", (-4.90536454209226,-0.6272588643672109), NE * labelscalefactor); dot((5.84,-0.72),dotstyle); label("$C$", (5.905425048021831,-0.5404949029698266), NE * labelscalefactor); dot((-3.65,3.09),linewidth(4pt) + dotstyle); label("$E$", (-3.586552328852017,3.225061021676652), NE * labelscalefactor); dot((0.43,-0.76),linewidth(4pt) + dotstyle); label("$D$", (0.49135385682504773,-0.6272588643672109), NE * labelscalefactor); dot((1.76,3.13),linewidth(4pt) + dotstyle); label("$F$", (1.827518862344767,3.2771193985150826), NE * labelscalefactor); dot((-3.9237156783181897,2.289433091234769),linewidth(4pt) + dotstyle); label("$M$", (-3.8468442130441702,2.4268325768207166), NE * labelscalefactor); dot((2.5996691485249728,2.3376651417105037),linewidth(4pt) + dotstyle); label("$N$", (2.6604528917596566,2.478890953659147), NE * labelscalefactor); dot((-0.6620232648966086,2.313549116472636),linewidth(4pt) + dotstyle); label("$O$", (-0.5845192645025182,2.4441853691001936), NE * labelscalefactor); dot((-4.3392876182874085,1.0739632818511171),linewidth(4pt) + dotstyle); label("$P$", (-4.263311227751615,1.2121371172573363), NE * labelscalefactor); dot((3.8745063779042277,1.1346937365364522),linewidth(4pt) + dotstyle); label("$R$", (3.944559520440945,1.2815482863752437), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Use reference triangle $\triangle{ABC}$ so we have \[A:= (1, 0, 0), \ B:= (0, 1, 0), \ C:= (0, 0, 1), \ D:= (0, \tfrac{1}{2}, \tfrac{1}{2}), \ E:= (\tfrac{1}{2}, \tfrac{1}{2}, 0), \ F:=(\tfrac{1}{2}, 0, \tfrac{1}{2}).\]From Stewart's Theorem we can easily compute $AD=\frac{\sqrt{2(b^2+c^2-a^2)}}{2}$. Let $k=2\cdot AD$. Using the angle bisector theorem we can easily compute $M:=(\tfrac{ac}{a+k}, \tfrac{ck}{a+k}, 0)$ and $N:=(\tfrac{ab}{a+k}, 0, \tfrac{bk}{a+k})$. Intersecting $MN$ and $AD$ yields $O:=(\tfrac{a}{a+k}, \tfrac{k}{2a+2k}, \tfrac{k}{2a+2k})$. From that we have $P:=(\tfrac{2a-k}{2a}, \tfrac{k}{2a}, 0)$ and $R:=(\tfrac{2a-k}{2a}, 0, \tfrac{k}{2a})$. The displacement vector $\vec{PR}$ is $<0, \tfrac{k}{2a}, -\tfrac{k}{2a}>,$ so we have \[PR^2=\frac{k^2}{4}=AD^2,\]as desired.

AM/MB = AD/DB = AD/DC = AN/NC so MN || BC. AP/MP . MO/NO . FN/FA = 1 = AR/NR . NO/MO . EM/EA so AP/MP = AR/NR so PR || MN. Note that AP/AB = PR/BC, BM/BD = AM/AD and FC/NC = AP/AM so AP.PR.AM / AM.BC.AD = FC.AP.BM / NC.AB.BD so PR/AD = 2(FC.BM / NC.AB) = 1/2 so PR/AD = 1 so PR = AD.