Ptolemy's Theorem

Examining the similar triangles in (c), we have ${\frac {j}{d}}={\frac {b}{p}}$, so $jp=bd$

Examining the similar triangles in (d), we have ${\frac {j}{c}}={\frac {a}{q}}$, so $jq=ac$

So $j(p+q)=bd+ac$ ... Q.E.D.