In the figure AB|

Given that ∠EGB=50°

Since, AB||CD and EF is a transversal,

∠GHD=∠EGB=50° (corresponding angles) ……….(1)

∠CHF=∠EGB=50° (alternate exterior angles)

∠AGH=∠GHD =50° (alternate interior angles)

Now, ∠BGH+∠GHD=180° (co-interior angles)

⇒ ∠BGH+50° =180° (from (1))

⇒ ∠BGH=180°-50°

⇒ ∠BGH=130°

∠GHC=∠BGH=130° (alternate interior angles)

∠AGE=∠GHC=130° (corresponding angles)

∠DHF=∠AGE=130° (alternate exterior angles)

Hence, ∠GHD=50°, ∠CHF=50°, ∠AGH=50°, ∠BGH=130°, ∠GHC=130°, ∠AGE=130° and ∠DHF=130°