Question

solution...............,......,...,................,..​

Answer 1

Answer:

there is two methods to solve this problem.. but both will give same answer..

1st method

join E to F

now EF ll BC

and the drawn line is bisector of angle AED and angle AFD

so

angle AEF = angle FED

and

angle AFE = angle EFD

this makes

angle AEF = X° = angle FED

and

angle AFE = y° = angle EFD

now in triangle

angle EDF = 180 - x° - y°

2nd method

here BE ll DF

thus angle x° = angle FDC

here ED ll FC

thus angle y° = angle EDB

now these angle form linear pair on line BC

thus

angle FDC + angle EDB + angle EDF = 180

x° + y° + angle EDF = 180

hence

angle EDF = 180 - x° - y°

hope this will help you