Given A,C are two points on DP and DQ of $\triangle DPQ$, B is the intercept of AQ and CP, PQ=AP+CQ, $\angle \alpha+\angle \beta=180^o$. If also given AP=CQ, it can be proven $\angle PDQ$ = 60$^o$ and $\angle PBQ$=120$^o$.
Question:
If the requirement of AP=CQ removed, are there any other conditions that can lead to the same conclusion $\angle PDQ$ = 60$^o$ and $\angle PBQ$=120$^o$?
It seems that we need to construct something to link PQ,AP and CQ together, need help to continue.
The suggested solution assumes some basic knowledge.
Let me know if this is what you have in mind and if all is clear.