The given figure shows a triangle $\mathrm{ABC}$ in which $\mathrm{AD}$ bisects angle $\mathrm{BAC}. \mathrm{EG}$ is perpendicular bisector of side $\mathrm{AB}$ which intersects $\mathrm{AD}$ at point $\mathrm{F}$. Prove that:

$\mathrm{F}$ is equidistant from $\mathrm{AB}$ and $\mathrm{AC}.$
 

Construct a triangle $\mathrm{ABC},$ with $\mathrm{AB} = 7 \mathrm{cm}, \mathrm{BC} = 8 \mathrm{cm}$ and $\angle \mathrm{ABC} = 60°.$ Locate by construction the point $\mathrm{P}$ such that:

$\mathrm{P}$ is equidistant from $\mathrm{B}$ and $\mathrm{C}.$ $\mathrm{P}$ is equidistant from $\mathrm{AB}$ and $\mathrm{BC}.$ Measure and record the length of $\mathrm{PB}.$