Given, a rectangle ATEF in which EF = EB. Then, ΔFEB is an isosceles triangle. Therefore, by the angle sum property of a triangle, we have
∠EFB+∠EBF+∠FEB=180∘ [angle sum property of triangle]
⇒∠EFB+∠EBF+90∘=180∘ [∵ in a rectangle, each angle is of 90∘]
⇒2∠EFB=90∘ [∵EB=EF⇒∠EFB=∠EBF]
∠EFB=45∘ and ∠EBF=45∘
Now, ∠x=180∘−45∘=135∘ [linear pair]
and ∠EFB+∠y=90∘ [∵ in a rectangle, each angle is of 90∘]
⇒∠y=90∘−45∘=45∘ .