We are given that AB || DE
and ∠BAC = 35° and ∠CDE = 53°
AB || DE (given)
⇒ ∠BAE = ∠AED = 35°
(alternate angles)
Now in ∆CDE,
∠CDE + ∠E + ∠DCE = 180°
(angle sum property of a triangle)
⇒ 53° + 35° + ∠DCE = 180°
⇒ ∠DCE = 180° – 88° = 92°
⇒ ∠DCE = 92°
Hence, ∠DCE = 92°.