ABC is any triangle and P is a point on the side BC. If Q is a point such that $\overline{\mathrm{PQ}} = \overline{\mathrm{AP}}+\overline{\mathrm{PB}} + \overline{\mathrm{PC}}$, then

$\overline{\mathrm{AQ}} = \overline{\mathrm{AB}} + \overline{\mathrm{AC}}-\overline{\mathrm{BC}}$

$\overline{\mathrm{AQ}} = \overline{\mathrm{AB}} + \overline{\mathrm{BC}} + \overline{\mathrm{AC}}$

Q depends on the position of P

Q is fixed for all position of P on BC