Question

show that the line segment which join the midpoint of the oblique sides of a trapezium is parallel to parallel sides.

Please give answer in explained and in copy work. ​

Answer 1

Step-by-step explanation:

Let the trapezium be ABCD with E and F as the mid Points of AD and BC, Respectively Produce AD and BC to Meet at P.

In ∆PAB, DC||AB

Applying Thales’ theorem, we get PD/DA = PC/CB Now, E and F are the midpoints of AD and BC, respectively.

PD/3DE=PD/2CF

PD/DE=PD/CF

Applying the converse of Thales’ theorem in ∆ PEF, we get that DC Hence, EF || AB. Thus. EF is parallel to both AB and DC.

This completes the proof.

Answer 2

Answer:

Let the trapezium be ABCD with E and F as the mid Points of AD and BC, Respectively Produce AD and BC to Meet at P.

Applying Thales’ theorem, we get PD/DA = PC/CB

Now, E and F are the midpoints of AD and BC, respectively.

Applying the converse of Thales’ theorem in ∆ PEF, we get that DC Hence, EF || AB.

Thus. EF is parallel to both AB and DC.

This completes the proof.