Proof:
On the diagonal BD
locate a point M such that angles ACB and MCD be equal.
Since angles BAC and BDC subtend the same arc,
they are equal. Therefore,
triangles ABC and DMC are similar. Thus we
get CD/MD = AC/AB,
or
AB·CD = AC·MD.
Now, angles BCM and ACD are also equal; so triangles BCM and ACD are similar.
Therefore, BC/BM = AC/AD, or BC·AD =
AC·BM.
Summing up the two identities we obtain
AB·CD + BC·AD = AC·MD + AC·BM = AC·BD