Question

3. In trapezium ABCD, AB II CD and E is the midpoint of AD. A line drawn through E and parallel to AB intersects BC at F Prove that F is the midpoint of BC and EF (AB + CD). ​

Answer 1

Answer:

Given ABCD is a trapezium.

We have to prove, F is the mid point of BC, i.e., BF=CF

Let EF intersect DB at G.

In ΔABD E is the mid point of AD and EG∣∣AB.

∴ G will be the mid-point of DB.

Now EF∣∣AB and AB∣∣CD

∴ EF∣∣CD

∴ In ΔBCD, GF∣∣CD

⇒ F is the mid point of BC.