Question

plzz solve this for me​

Answer 1

Step-by-step explanation:

AD = DB ---> A-D-B

AE =EC ---->A-C-C

now in ∆AED and ∆CEF,

angle AED = angle CEF ( vertically opposite angle)

angle DAE = angle ECF ( alternate angle )

∆AED. = ∆ CEF ( by AA test )

DE = EF (by c.s.c.t.)------->1

2DE =DF ----------->2

In quadrilateral DBCF

DE || BC. ( given)

DF || BC

DE = 1/2 BC ( by midpoint theroam)

BC = 2DE

BC= DF ( from 2)

opposite sides of quadrilateral are equal

therefore it is a parallelogram

from above

DB || FC

DE = FC