Find the value of $x$ in the triangle below For reference: Given the triangle below, where $AB = BC$, find the value of $x$ (Answer: $150^o$)

My progress:

$2\alpha+a+2\theta = 90^o\implies (\alpha+\theta) = 45^o - \frac{a}{2}\\
\alpha+b + 2\theta = 90^o\\\
\alpha+b = a+2\alpha \implies \alpha = b-a\\
x=180^0-(\alpha+\theta)=135^o +\frac{a}{2}\\
180-(3\theta+2\alpha)+180-(a+b)+x = 360 \implies x = 3\theta+2\alpha+a+b\\
$
Some detail is missing

It is not possible to change the solution and get the original figure
 A: (I use $t,a$ instead of $\theta$, $\alpha$ for ease of typing.)

In the picture, $E$ is the mid point of $AC$. A prime denotes reflection w.r.t. $BE$, so for instance $A'=C$. Build $K'$, then $L=AK\cap A'K'\in BE$. Divide the $4t$ angle in $\hat B$ in four $t$-parts, the angles $\widehat{KAB}$, $\widehat{K'CB}$ in two $a$-parts. Draw all remaining angle bisectors in $\Delta BAL$ and $\Delta BCL$. Denote by $u$ the half angles built in $L$ in these triangles. The sum of angles in them is $180^\circ$, so
$$
a+t+u=\frac 12180^\circ=90^\circ\ .
$$
Since $KK'\|AA'=AC$ is perpendicular to $BLE$, we have two $(a+t)$-angles in $\Delta LKK'$, they are then copied in $\Delta LAC$.
The sum of angles in $\Delta ABC$ is $180^\circ$ so:
$$
180^\circ=4t+(2a+(a+t))+(2a+(a+t))=6(a+t)\ .
$$
This gives $a+t=30^\circ$, so the angle in $K$ in $\Delta BKC$ is the remaining amount of $180^\circ$.
$\square$
A: $\angle PBC = \frac{\angle ABC}{2} =2θ\\
 \angle PCB= 2\alpha ( by ~symmetry)\\
 \triangle  BPC : K (incenter) \implies  \angle CPK= \angle KPB(I)\\
 \angle APH= \angle KPB(o.b.v.)(II)\\
\angle APH= \angle HPC(III)\\
 (De (I),(II)e(III)\angle KPB=\angle CPK= \angle HPC
\therefore \angle KPB=\angle CPK=\angle HPC=60^o\\
\triangle  BPC: 2θ+2α+120^o=180^o \implies θ+α=30\\
\triangle BKC: x=\angle BKC=180^o−(θ+α)=150^o$

