∵ AB || CD and transversal line intersect E and D respectively.
∴ ∠AED = ∠CDP (corresponding angles)
=> ∠AED = 34°
Ray EF is inclined on line AB at point E
∴ ∠AEF + ∠BEF = 180°
=> (∠AEP + ∠PEF) + ∠BEF = 180° [ ∵ ∠AEF = ∠AEP + ∠PEF]
=> 34° + ∠PEF + 78° = 180°
=> ∠PEF = 180°- 112° = 68°
=> ∠DEF = 68° …(1)
∵ EF || DQ and transversal line DE intersects at E and D respectively.
∵ ∠FED = ∠PDQ = 68° [ ∵ ∠FED = ∠PEF = 68°]
=> ∠PDQ = 68°
So ∠PDQ = ∠DEF = 68°, ∠AED = 34°
