For reference:
Through the center of a triangle $ABC$ is drawn a parallel to $AC$ that intercepts at $P$ to $AB$ and in $Q$ to $BC$. Calculate $PQ$ if $\frac{1}{k}+\frac{1}{b} = 0.25$, $AC = b$ and $k$ is the perimeter of triangle $PBQ$ (Answer:$ 4$)
My progress: Here's my figure and the relationships I found:
$\displaystyle \small \triangle BPQ \sim \triangle BAC \implies \frac{BP}{BA} =\frac{BQ}{BC} =\frac{PQ}{b}=\frac{k}{2p_{BAC}}$
$\displaystyle \small 0.25kb=b + k\implies kb = 4(b+k)$
$\displaystyle \small PQ = \frac{4(b+k)}{2p_{BAC}}...$
$\angle PDA = \angle DAC = A/2 =\angle PAD$. So $\triangle APD$ is isosceles. Hence $PA=PD$. Similarly, $QD=QC$.
So perimeter of $\triangle BPQ$ is $$K = BP+PD+DQ+QB $$$$= BP+PA+QC+BQ=BA+BC$$
Hence proceeding from your solution $$\frac{PQ}{b}=\frac{K}{b+BA+BC}=\frac{K}{b+K}$$ $$\Rightarrow \frac{1}{PQ}=\frac{1}{b}+\frac{1}{K}$$