Question

Prove that if midpoints of non-parallel sides of a trapezium are joined, this line is parallel to parallel sides of the trapezium?

Answer 1

For proof see below.

Explanation:

Basic Proportionality Theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.

Reverse is also true that is, if a line divides the two sides of a triangle in the same ratio, then it is parallel to the third side.

Now let us revert to the problem. Let `ABDC` be the trapezium with parallel sides `AB` and `CD` and non-parallel sides `AC` and `BD`, whose mid-points are `E` and `F` respectively. `EF` are joined. We have to prove that `EF`||`CD`.

For this let us join `AF` and produce it to meet extended `CD` at `G`.

Now in `Deltas` `ABF` and `DGF`, we have

`BF=DF` (as `F` is midpoint of `BD`)

`/_ABF=/_FDG` (alternate opposite angles) and

`/_BFA=/_DFG` (vertically opposite angles)

Therefore `DeltaABF-=DeltaDGF` (ASA)

Hence, `AF=FG` and in `DeltaACG`,

as `AE=EC` and `AF=FG` i.e. `(AE)/(EC)=(AF)/(FG)`

and hence `EF`||`CD`||`AB`

Quod Erat Demonstrandum