The correct option is A True
Given: D is mid point of AB and E is mid point of BC, F is any point on AC and EF∥AB
Now, in △ABC,
E is mid point of BC and EF∥AB
By Mid point Theorem, F is mid point of AC
Also, D is mid point of AB and F is mid point of AC
Hence, by mid point theorem, DF∥BE
Since, DF∥BE and EF∥ABorBD
Hence, BEFD is parallelogram.