<s>[INST] Calculate the measure of segment AB in triangle rectangle ABC For reference:
Given the triangle $ABC$, straight at $B$. The perpendicular bisector of $AC$ intersects at $P$
with the angle bisector of the outer angle $B$, then $AF \parallel BP$  ($F\in BC$) is drawn.
If $FC$ = $a$, calculate BP(x). (Answer: $\frac{a\sqrt2}{2})$

My progress:

Point P is on the circumcircle of ABC because the angle $\measuredangle ABP = 135^o$
$ Where~ AC = 2R\\
\triangle CBP \rightarrow
PC^2 = BP^2 + BC^2 - \sqrt2BCBP\\
but~PC = PA = R\sqrt{2} \text{(since P is in the bisector AC)}\\
2R^2 = BP^2 +BC^2 - \sqrt{2}BCBP\\
0= BP^2 - \sqrt{2}BCBP + BC^2 - 2R^2\\
0= (BP - \frac1{ \sqrt{2}}BC)^2 + \frac{BC^2}2 - 2R^2\\
0= (BP - \frac1{ \sqrt{2}}BC)^2 + \frac{BC^2-4R^2}2\\
0= (BP - \frac1{ \sqrt{2}}BC)^2 - \frac{AB^2}2\\
$
I can't find the relationship between BC, AB and a...
If anyone finds another way to solve by geometry I would be grateful
 [/INST] 
You have done most of the work by coming up with the nice construction. I will just mark point $Q$, draw chord $CQ$ and use it to find the answer.
As $BP$ and $AQ$ are parallel, $\angle AFB = \angle FBP = 45^0$
So $\angle CFQ = 45^0$ and as $\angle CQF = 90^0$, $CQ = \cfrac{a}{\sqrt2}$
But also note that, $\angle CAQ = \angle BAP = \angle A - 45^0$
So we must have, chord $BP$ = chord $CQ$.
$\therefore x = \cfrac{a}{\sqrt2}$
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