How to prove that two lines of a shape are parallel? ∠A=∠B, and P is a point on AB such that AD=PD. Prove that DP is parallel to CB. Please state how also ($SSS$, $ASA$, $SAS$, $CPCTC$, Isosceles Biconditional, interior angles, etc.)

 A: Using Isosceles Triangle Theorem $\overline{AD} \cong \overline{PD} \Rightarrow \angle A \cong \angle APD$. Consider $\overleftrightarrow{DP}$ and $\overleftrightarrow{CB}$. Consider the line $\overleftrightarrow{AB}$ which is a transversal to $\overleftrightarrow{DP}$ and $\overleftrightarrow{CB}$.
Using the converse of Alternate Interior Angle Theorem we see that $\angle APD \cong  \angle ABC \Rightarrow \overleftrightarrow{DP}\parallel \overleftrightarrow{CB}$.
Isosceles Triangle Theorem
Let $\triangle ABC$ be a triangle such that $\overline{AC} \cong \overline{BC}$. Then $\angle A \cong \angle B$.
Proof:
Consider the correspondence $A-B-C\leftrightarrow B-A-C$, from which we can deduce that $\overline{AC} \cong \overline{BC}$, $\angle C \cong \angle C$ and $\overline{BC} \cong \overline{AC}$. By SAS Theorem $\triangle ABC \cong \triangle BAC$, therefore, because of "parts of congruent triangles are congruent" $\angle A \cong \angle B$.
Converse of Alternate Interior Angle Theorem
Consider the figure shown in the problem as a general case. If $\angle APD \cong \angle ABC$, then $\overleftrightarrow{DP} \parallel\overleftrightarrow{CB}$.
Proof:
Suppose that $\overleftrightarrow{DP} \nparallel\overleftrightarrow{CB}$. Therefore they will intersect at a point $M$. Now consider the triangle $\triangle PMB$. By the external angle theorem $\angle APD >\angle ABC \rightarrow\leftarrow.$ Therefore  $\overleftrightarrow{DP} \parallel\overleftrightarrow{CB}$.
BTW, the question must be "...such that AD = PD..." Here is a graphical counterexample of the condition you posted.

A: $AD=DP$ implies $\angle(DPA)=\angle(DAP)$. By assumption, the latter equals $\angle(ABC)$ so we have $\angle(DPA)=\angle(ABC)$.
