Let $M=BQ\cap AP$ and $N=DP\cap CQ$. Claim 1: $MN\parallel AD$ Proof: $$\dfrac{MQ}{MB}=\dfrac{PQ}{AB}=\dfrac{PQ}{CD}=\dfrac{QN}{NC}$$meaning $MN\parallel AD$ $\square$ Claim 2: The quadrilateral $MPQN$ is cyclic Proof: Let $\angle A=\angle C=2\alpha$. Then $\angle MPQ=\alpha$ since $PQ\parallel AB$ from this $\angle PQN=\alpha$. Also, $\angle PMN=\angle PAD=\alpha$ implying $\angle PMN=\angle PQN$ or $MPQN$ is cyclic $\square$ Finishing: $\angle ABQ=\angle BQP=\angle MNP=\angle ADP$ $\blacksquare$

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Bexultan wrote: Let $M=BQ\cap AP$ and $N=DP\cap CQ$. Claim 1: $MN\parallel AD$ Proof: $$\dfrac{MQ}{MB}=\dfrac{PQ}{AB}=\dfrac{PQ}{CD}=\dfrac{QN}{NC}$$meaning $MN\parallel AD$ $\square$ Claim 2: The quadrilateral $MPQN$ is cyclic Proof: Let $\angle A=\angle C=2\alpha$. Then $\angle MPQ=\alpha$ since $PQ\parallel AB$ from this $\angle PQN=\alpha$. Also, $\angle PMN=\angle PAD=\alpha$ implying $\angle PMN=\angle PQN$ or $MPQN$ is cyclic $\square$ Finishing: $\angle ABQ=\angle BQP=\angle MNP=\angle ADP$ $\blacksquare$ nice solution. where did you draw diagram?

geogebra classic