Point $P$ lies in the interior of $\Delta ABC$. Prove that $PB + PC < AB + AC$.

(N/A) $1$. Extend $BP$ to intersect $AC$ at point $D$.
$2$. In $\Delta ABD$,by the triangle inequality theorem,the sum of two sides is greater than the third side: $AB + AD > BD$.
$3$. Since $BD = BP + PD$,we can write: $AB + AD > BP + PD$ --- (Equation $1$).
$4$. In $\Delta PDC$,by the triangle inequality theorem: $PD + DC > PC$ --- (Equation $2$).
$5$. Adding Equation $1$ and Equation $2$: $(AB + AD) + (PD + DC) > (BP + PD) + PC$.
$6$. Simplifying the inequality: $AB + (AD + DC) + PD > BP + PD + PC$.
$7$. Since $AD + DC = AC$,we get: $AB + AC + PD > BP + PC + PD$.
$8$. Subtracting $PD$ from both sides,we obtain: $AB + AC > PB + PC$,which is equivalent to $PB + PC < AB + AC$.