Straightforward with barycentric coordinates

here is an error, in conclusion, the right thing is ΒΡ=ΡF

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yes there was a typo, the ratio of the lengths is wanted and not their equality

Is the answer $4$.

It is easy. Just using Menelaus Theorem many times we would find the result

my solution is right, just made an accounting error, 1/3 instead of rectal 2/3: $\begin{array}{l} \frac{{{\rm{BP}}}}{{{\rm{PQ}}}} = \frac{{{\rm{PF}}}}{{{\rm{PQ}}}} = \frac{{{\rm{PQ + QF}}}}{{{\rm{PQ}}}} = 1 + \frac{{{\rm{QF}}}}{{{\rm{QP}}}}\mathop {}\limits_{}^{} {\rm{ }}(1)\\ \frac{{{\rm{QF}}}}{{{\rm{PQ}}}} = \frac{{{\rm M}{\rm N}}}{{{\rm{P}}{\rm Z}}} = \frac{{\frac{{{\rm{DE}}}}{2}}}{{\frac{{{\rm M}{\rm N} + {\rm{DE}}}}{2}}} = \frac{2}{3}\mathop {}\limits_{}^{} \mathop \Rightarrow \limits^{(1)} \mathop {}\limits_{}^{} \frac{{{\rm{BP}}}}{{{\rm{PQ}}}} = \frac{5}{3} \end{array}$

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I guess the correct answer for $\boxed {BP:PQ=5:3}$ .......that's what the official solution atleast says...... Ok so here's the solution using Barycentric Coordinates, It is trivial to find out that $P\equiv (x:2:1) $ and since, $P $ lies on $BF $, where $F \equiv (1:0:1) $ it follows that, $$\boxed {P \equiv (1:2:1)} $$ Similarly, we get that $Q \equiv (2:y:2) $ and $ Q \equiv (x:1:2) $ $$\implies \boxed{Q \equiv (2:1:2)} $$ Let WLOG, assume that $[ABC]=1 $, then from Barycentric Coordinates for $P $ and $Q $, we have, $$[ABP]=\frac {1}{4} \text { and} [ABQ]=[ABP]+[APQ] = \frac {2}{5} $$Hence, it's easy to notice that $$\frac {[ABQ]}{[ABP]} = \frac {[APQ]}{[ABP]} +1 = \frac {8}{5} \implies \frac {[APQ]}{[ABP]} =\frac {PQ}{BP} = \frac {3}{5} $$ Hence, $$\boxed {BP:PQ=5:3} $$

Menelaos in $\triangle ADC$ with transversal $\overline {BPF}$ gives $AP=3PD, PD$ is midline of $\triangle BEF\implies EF=2PD$, so $\frac{AP}{EF}=\frac{3}2=\frac{PQ}{QF}$, last relation being equivalent to $\frac{PQ}{PF}=\frac{3}5$, but $PF=BP$, hence $\frac{BP}{PQ}=\frac{5}3$. Best regards, sunken rock

easy solution using midpoint theorm and similar traingles but for that we need to do construction i made attachment once check it for solution

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