In quadrilateral ABCD, ∠ABC and ∠ADC are complementary, and BE and DF bisect the internal angles of quadrilateral ABCD ∠ABC and ∠ADC respectively.

So ∠ Abe+∠ ADF = 90.

If BE is parallel to DF

Then ∠ADF=∠AEB

So ∠ Abe +∠ AEB = 90.

So ∠ a = 90.

Hope to adopt