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
 A: $BP \parallel AF \implies ∠PBF= ∠BFA=45° \text{  (alternate interior angles)}$
$ \implies ∠BAF=∠BFA  \text{  (sum of angles in triangle)} $
$\implies AB=FB  \text{ (sides opposite to equal angles in a triangle)}$
$\therefore BC= CF+FB=a+AB \implies AB=BC-a$
Using this, in the equation you got
$(BP - \frac1{ \sqrt{2}}BC)^2 - \frac{AB^2}2 =0 \implies  (BP - \frac1{ \sqrt{2}}BC)^2 - (\frac{BC-a}{\sqrt{2}})^2=0$
$\implies \{(BP - \frac1{ \sqrt{2}}BC)+(\frac{BC-a}{\sqrt{2}})\}\{(BP - \frac1{ \sqrt{2}}BC)-(\frac{BC-a}{\sqrt{2}})\}=0$
$\implies (BP - \frac{a}{ \sqrt{2}})(BP - \frac2{ \sqrt{2}}BC+\frac{a}{ \sqrt{2}})=0$
$\implies BP=\frac{a}{ \sqrt{2}} \text{  or   }BP=\sqrt{2} BC-\frac{a}{ \sqrt{2}} \tag{i} \label{i}$
Let us assume $BP=\sqrt{2} BC-\frac{a}{ \sqrt{2}} \tag{ii}$
$$\text{In  }\triangle ABF, AB+BF>AF \implies 2(BC-a)>AF (\because AB=BF=BC-a$$
$$\implies 2BC-2a+a>AF+a>AC (\because AF+a>AC \text{  in  } \triangle AFC$$
$$\implies \sqrt{2} BC-\frac{a}{ \sqrt{2}}>\frac{AC}{\sqrt{2}} \text{ 
 (divided both sides by }\sqrt{2})$$
$$\implies BP>\frac{AC}{\sqrt{2}}=\frac{2R}{\sqrt{2}}=\sqrt{2}R\text{ (using (ii))}$$
$\implies BP>AP \implies \overset\frown{BP}>\overset\frown{AP}$ which is a contradiction. Therefore our assumption (ii) is wrong.
$\therefore BP\neq \sqrt{2} BC-\frac{a}{ \sqrt{2}}$ and (i) $\implies BP =\frac{a}{ \sqrt{2}} $
$\textbf{Method 2: Using similarity of triangles}$
$\begin{array}{l}
\text{In  } \triangle ABP \text{  and  } \triangle AFC\\
\angle ABP= \angle AFC =135° (\because \angle AFC=180°-\angle BFA)\\
\angle BAP=\angle FAC (\because \angle BAP=\angle BAC - \angle PAC =\angle BAC - 45°=\angle BAC - \angle BAF=\angle FAC)
\end{array}$
$\therefore \triangle ABP \sim \triangle AFC$ by AAA similarity criterion.
$\implies \frac{BP}{FC}=\frac{AP}{AC} \implies  \frac{BP}{a}=\frac{\sqrt{2} R}{2R}=\frac1{\sqrt{2}}$
$\therefore BP=\frac{a}{\sqrt{2}}$
A: 
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 $CQ$.
$\therefore x = \cfrac{a}{\sqrt2}$
A: An uglier brute-force method by some trigonometric identities:

Let the green marked $\angle BCA$ be $\theta$. Then the corresponding angle at centre $\angle BEA = 2\theta$, and $\angle PEB = 90^\circ - 2\theta$.
By the laws of cosine,
$$\begin{align*}
BP^2 &= EB^2 + EP^2 - 2 EB\cdot EP \cos \angle PEB\\
&= 2R^2 - 2 R^2 \cos (90^\circ - 2\theta)\\
&= 2R^2 - 2R^2\sin 2\theta\\
&= 2R^2\left(\sin^2\theta -2\sin\theta\cos\theta +\cos^2\theta\right)\\
&= \frac12 (2R)^2\left(\cos\theta-\sin\theta\right)^2\\
&= \frac12 \left[(BF+a)-AB\right]^2\\
&= \frac{a^2}{2} & (AB=BF)\\
x = BP &= \frac{a}{\sqrt2}
\end{align*}$$
