Proving that in a quadrilateral $ABCD$, where $P$ is the intersection of angle bisectors of $BAD$ and $ADC$, $|BP|=|CP|$

Question

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.

Answer 1

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}

Answer 2

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.

Answer 3

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 $BP=CP$