E and F are respectively the midpoints of non parallel sides AB and BC of a trapezium ABCD, prove that EF || AB.
Answers
- Let ABCD be a trapezium in AB || DC. Let E and F be the midpoints of AD and BC respectively E and F are joined.
- We have to show that EF || AB.
- If possible,let EF be not parallel to AB then draw EG || AB, meeting BC to G.
- Now, AB || EG || DC and the transversal AD cuts them at A,E,D respectively such that AE = ED.
Also, BGC is the other transversal cutting AB,EG and DC at B, G and C respectively.
∴ BG = GC (by intercept theorem).
This shows that G is the midpoint of BC.
Hence,G must coincide with F [∴ F is the midpoint of BC].
Thus,our supposition is wrong.
Hence, EF || AB.
E and F are respectively the midpoints of non parallel sides AB and BC of a trapezium ABCD, prove that EF || AB.
Let ABCD be a trapezium in AB || DC. Let E and F be the midpoints of AD and BC respectively E and F are joined.
We have to show that EF || AB.
If possible,let EF be not parallel to AB then draw EG || AB, meeting BC to G.
Now, AB || EG || DC and the transversal AD cuts them at A,E,D respectively such that AE = ED.
Also, BGC is the other transversal cutting AB,EG and DC at B, G and C respectively.
∴ BG = GC (by intercept theorem).
This shows that G is the midpoint of BC.
Hence,G must coincide with F [∴ F is the midpoint of BC].
Thus,our supposition is wrong.
Hence, EF || AB.