ComplexPhi wrote: The lines of support of the internal bisectrices of the angles $CAD$ and $CBD$ meet at $P$ what do you mean by this? please elaborate? do you mean angle bisectors??

Yes, I mean the lines of support of the internal angle bisectors of $CAD$ and $CBD$ meet at $P$ .

As $\triangle ABC$ and $\triangle ABD$ have equal perimeter, then $CA+CB=DA+DB$ $\Longrightarrow$ $D$ is on the ellipse $\mathcal{E}$ with foci $A,B$ that passes through $C.$ If the tangents of $\mathcal{E}$ at $C,D$ meet at $P^*,$ then by conic tangent property, $AP^*$ and $BP^*$ bisect $\angle CAD$ and $\angle CBD$ $\Longrightarrow$ $P \equiv P^*$ $\Longrightarrow$ $PA,PB$ are isogonals WRT $\angle CPD,$ i.e. $\angle APC=\angle BPD.$

The condition means that $C,D$ lie on an ellipse $\epsilon$ with foci $A,B$ and the bisectrix condition me as that $P$ is the point of intersection of tangents to $\epsilon$ at $C,D$. The isogonal condition follows as well. All these are standard results, not a very good TST example.

I would like to provide a solution without ellipse: Take the points $F$ on the line $(AD$ and $E$ on the line $(BC$ such that $DB=DF$ and $CA=CE$. The condition implies $AF=BE$. Take $P'$ to be the intersection of the perpendicular bisectors of $BF$ and $AE$. We will prove that $P,P'$ are the same. The perpendicular bisector condition implies $AP'=P'E$,$BP'=P'F$, and $\angle P'EB=\angle P'AC$. Since $AF=BE$ we have $\triangle P'AF=\triangle P'EB$ so $\angle FAP'=\angle P'EB=\angle P'AC$ so $AP'$ is the angle bisector of $DAC$. In the same way we have $BP'$ is the angle bisector of $DBC$, so $P=P'$. Since $\angle APF=\angle BPE$ and $APC=CPE$, $\angle DPF=\angle DPB$, the conclusion follows.