Translate $\Delta APB$ $AD$ units in the direction antiparallel of $\overrightarrow{AD}$ to $\Delta A'P'D'$. Note that $\angle AP'D=\angle CPD$, so quadrilateral $APBP'$ is cyclic. Also $AP'=DP$, $BP'=CP$, and $PP'=AD$ Now we make those substitutions in our equation to be proven We want to prove that $AB\cdot PP'=BP\cdot AP'+AP\cdot BP'$. This is true by ptolemy's theorem