ABCD is a cyclic quadrilateral whose diagonals intersect at P.  If AB = BC, $$\angle$$DBC = $$70^\circ$$ and $$\angle$$BAC = $$30^\circ$$, then the measure of $$\angle$$PCD is:

As per the given question,

From the given diagram we can see that in triangle $$\triangle BAC$$ and $$\triangle BDC$$ both are on the same base and in the same circle,

So, $$\angle BAC$$ =\angle BDC=30$$

Now, in $$\triangle BAC$$, BA=BC (given in the question)

So, $$\angle BAC=\angle BCA=30$$

Now, in $$\triangle BDC$$

$$\angle DBC +\angle BDC+\angle BCD=180$$

So, $$\angle BCD=180-70-30=80^\circ$$

Now, $$\angle BCD=80^\circ$$

So, $$\angle PCD=80-\angle PCB=80-30=50^\circ$$