Alternate angles $ABD$ and $CDB$ are equal (Fig. 2), hence $\angle BCA + \angle ABD + \angle BDA + \angle ACD = 180^\circ$. The equality $\angle BCA + \angle ABD = \angle BDA + \angle ACD$ thus holds if and only if
$$\angle BCA + \angle ABD = 90^\circ.$$
(1)

Fig. 2
Points $K$ and $L$ lie on a circle with diameter $BD$. Hence the inscribed angles $BDK$ and $BLK$ are equal and (due to equal alternate angles $ABD$ and $CDB$)
$$\angle BLK + \angle ABD = \angle BDK + \angle CDB = 90^\circ.$$
Lines $KL$ and $AC$ are parallel if and only if $\angle BLK = \angle BCA$ which is by the last equality equivalent to (1). The equivalence is thus proven.