Proving that in a quadrilateral $ABCD$, where $P$ is the intersection of angle bisectors of $BAD$ and $ADC$, $|BP|=|CP|$ Given a quadrilateral $ABCD$, where $|AD|=|AB|+|CD|$ the angle bisectors of $BAD$ and $ADC$ intersect at $P$. The task is to prove that $|BP|=|CP|$.
Since $|AD|=|AB|+|CD|$, the quadrilateral cannot be tangential, and thus the intersection points of adjacent angle bisectors are not concurrent and instead they form the vertices of a cyclic quadrilateral. This seems like an important piece of information, but I haven't been able to get any further.
Thank you for your help.
 A: Let $E$ be the point between $B$ and $D$ such that $AE = AB$ and $ED = CD$.  The point $P$ lies on the angle bisector of $\angle ADC$, which is also the perpendicular bisector of $\overline{EC}$.  Therefore, $CP = EP$.  Repeat this for the other angle to conclude that $BP = EP$.
Note: I am using $XY$ to denote distance and $\overline{XY}$ to denote the line segment.
A: 
Let us take advantage of |AD|=|AB|+|CD| by extending AB to F, DC to E, so that $BF=CD, CE=AB,  \color{orange}{AF= ED=AD}$.
Now, we have 3 sets of congruent triangles to play with.

$\triangle DPA\cong \triangle DPE (\color{orange}{AD=ED}, \alpha =\alpha',PD=PD)$, $\color {blue}{PA=PE}$.
$\triangle PAF\cong \triangle PAD (\color {blue}{PA=PE}, \beta =\beta',\color{orange}{AF=AD}), \color {green}{PF=PD},
\alpha=\alpha''$.
$\triangle PBF\cong \triangle PCD (\color {green}{PF=PD}, \alpha =\alpha'',BF=CD)$

\begin{align}\boxed {PB=PC}\end{align}
A: 
Let point $E$ be on $AD$ between $A$ and $D$ such that $AE=AB$. This automatically fixes $DE=CD$.Draw $BE$ and $CE$.
$\triangle ABE$ is isosceles and $AP$ bisects $\angle BAE$, hence $AP$ is the perpendicular bisector of $BE$. Similarly, it can be concluded that $PD$ is the perpendicular bisector of $CE$.
In $\triangle BCE$, the perpendicular bisector of $BE$ and $CE$ meet at point $P$. Hence, $P$ is the circumcentre of $\triangle BCE$ and thereafter $BP=CP$
