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In an ABC triangle, the cevianas AD, BE and CF compete in P. Show that

$\frac{S_{DEF}}{2S_{ABC}}=\frac{PD .PE.PF}{PA.PB.PC}$ Using areas relation, I found

$\frac{3S_{ABC}-\overbrace{(S_{PAB}+S_{PAC}+S_{PBC})}^{S_{ABC}}}{S_{ABC}}=\frac{PA}{PD}+\frac{PB}{PE}+\frac{PC}{PF}\implies 2S_{ABC}=S_{ABC}\left(\frac{PA}{PD}+\frac{PB}{PE}+\frac{PC}{PF}\right)\\ \frac{S_{PDF}}{S_{PAC}}=\frac{PD×PF}{PA×PC}\\ \frac{S_{PDF}}{S_{PAB}}=\frac{PD×PE}{PA×PB}\\ \frac{S_{PEF}}{S_{PBC}}=\frac{PE×PF}{PB×PC}\\ S_{PDF}+S_{PED}+S_{PFE}=S_{DEF}$ Can someone help me to finish this proof? Thanks for antetion!

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Let $a=\frac{a_1}{a_2}$, $b=\frac{b_1}{b_2}$, $c=\frac{c_1}{c_2}$. Per the Ceva's theorem, $abc = 1$. Evaluate

\begin{align} \frac{AP}{PD} &= \frac{S_{ABE}}{S_{DBE}} = \frac{\frac{b_2}{b_1+b_2}S_{ABC} }{ \frac{a_1}{a_1+a_2}\frac{b_1}{b_1+b_2}S_{ABC} } = \frac1b(1 + \frac1a )=c(1+a)\\ \end{align}

Likewise, $\frac{BP}{PE} = a(1+b)$ and $\frac{CP}{PF} = b(1+c)$. Then,

$$\frac{PA}{PD}\frac{PB}{PE}\frac{PC}{PF} =(1+a)(1+b)(1+c)\tag1 $$

Also, $$\frac{S_{AEF}}{S_{ABC}} = \frac{b_2}{b_1+b_2}\frac{c_1}{c_1+c_2} =\frac c{(1+c)(1+b)} \\ \frac{S_{BDF}}{S_{ABC}}= \frac a{(1+a)(1+c)} ,\>\>\>\frac{S_{CDE}}{S_{ABC}}= \frac b{(1+b)(1+a)} $$ and \begin{align} \frac{S_{DEF}}{S_{ABC}} & = \frac{S_{ABC}- S_{AEF}- S_{BDF} - S_{CDE}}{S_{ABC}} \\ &= 1-\frac c{(1+c)(1+b)} - \frac a{(1+a)(1+c)} - \frac b{(1+b)(1+a)}\\ &= \frac {abc+1}{(1+a)(1+b)(1+c)}\tag2\\ \end{align}

With $abc=1$, (1) and (2) lead to $$\frac{S_{DEF}}{2S_{ABC}}=\frac{PD}{PA}\frac{PE}{PB}\frac{PF}{PC}$$

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