Given AB||CD, ∠AEF = P°, ∠EFG = q°, ∠FGD = r°

Draw a line FH||AB||CD

∠HFG = ∠FGD = r° [Because HF||CD and alternate angles] ___________ (i)

∠EFH = ∠EFG - ∠HFG

⇒ ∠EFH = q – r ______________________ (i)

∠AEF + ∠EFH = 180° [Because AB||HF]

⇒ ∠AEF + ∠EFH = 180°

⇒ p + (q – r) = 180°

⇒ p + q – r = 180°Proved.