Just one point the point of intersection of diagnols?????!!!

Rushil wrote: Just one point the point of intersection of diagnols?????!!! I think it should be all points on the perpendicular bisector of AD and BC

shobber wrote: Rushil wrote: Just one point the point of intersection of diagnols?????!!! I think it should be all points on the perpendicular bisector of AD and BC Indeed, just draw those p.b. to divide ABCD into four regions, and assuming any other solution, prove that it leads to a contradiction in every one of those 4 cases (depending on where P lies in the 4 smaller rectangles)

Yes. Assume $P$ is not on the line, then $PA<PB$ and $PD<PC$. Then $\angle{PAD}<\angle{PBC}$ and $\angle{PDA}<\angle{PCB}$. Hence $\angle{DPA}>\angle{CPB}$. Cheers!

err...i think it's rather: By symmetry, we only have to consider one of the four cases. Assume P lies to the left of the p.b. of AB, and below the p.b. of AD, then AP > DP, then <PDC < <PAB. Similarly, BP > CP, so <PCD < <ABP. So overall, <DPC > <APB. Done.

Generalization. Let ABCD be a parallelogram. Ascertain the geometrical locus of the point P for which $\widehat{APB}\equiv\widehat{CPD}.$ Solution (own). I note $I\in AC\cap BD$. For the any point P, from the plane of the $\triangle ABC$, with the property $\widehat{APB}\equiv\widehat{CPD}$, the circumcircles $C_1(O_1,R_1),C_2(O_2,R_2)$ of the triangles APB, CPD have the same radius $R_1=R_2,\ O_1 O_2\parallel AD,\ IO_1=IO_2$ and the point I belongs to their common cord which is perpendicularly on AD. Thus, the geometrical locus of the point P is the line which pass through the point I and is perpendicularly on the line AD. Remark. Come on to extend at a convex quadrilateral ABCD !