Let's look for case when $R$ is on the arc $AB$ not containing by $O_2$ and $T$ is on the arc $AB$ not containing by $O_1$ Firstly $\angle CRE=\angle ARP=\angle ABT=180-\angle RCT$ $\Longrightarrow RE\Vert FT$ Similarly $RF\Vert ET$ $\Longrightarrow FRET$ is a paralelogram. $\angle QFT+\angle QBT=180-\angle FRE+\angle QBT=\angle PBQ+\angle QBT=180$$\Longrightarrow QFTB$ is cyclic. Similarly $RBCF,RADE$ and $PATE$ are cyclic. Using all these circles we get $\angle RCD=\angle ACD=\angle ATD=\angle ATE$ $\angle CRD=\angle ARD=\angle AED=\angle AET \Longrightarrow \bigtriangleup RCD \sim \bigtriangleup ETA$ $(1)$ Similarly $\angle BFR=\angle RCB=\angle ACB=\angle ATB=\angle QTP$ $\angle BRF=\angle BCT=\angle BAT=\angle QPT \Longrightarrow \bigtriangleup BFR \sim \bigtriangleup QTP$ $(2)$ From $(1)$ and $(2)$ $\frac{RD}{AE}=\frac{RC}{ET}$ and $\frac{BF}{QT}=\frac{FR}{TP}$ Multiplying side by side and usinf fact $RF=TE$ we get $BF.RD.TP=AE.QT.RC$ $(3)$ And from the power of point we have $QT.TA=TB.TP$ $(4)$ $RC.RA=RD.RB$ $(5)$ Multiplying $(3),(4),(5)$ we get $AE.TB.RB=BF.TA.RA$ I think other cases are similar to this one.

My solution is very similar to cefer's one, in fact the first few steps are just the same; but I finished it off in some other way. So, let me post a solution I got some while ago.

Sorry for posting a similar solution after $5$ years.