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If E and F are the two mid-points of the non-parallel sides of a trapezium ABCD respectively,then show that EF=1/2(AB+DC).


Let  ABCD  be a trapezoid with the bases  AB  and  DC  and the mid-line  EF.  Let us draw the straight line  DF  through the points  D  and  F
till the intersection with the extension of the straight line  AB  at the point G. Compare the triangles  DFC  and  FBG.

The segments  FC  and  BF  are congruent since the point  F  is the midpoint of the side  BC.  The angles  DFC  and  BFG are congruent as the vertical angles. The angles  DCF  and  FBG are congruent as the alternate exterior angles at the parallel lines  AB  and  DC  and the transverse  BC. Hence, the triangles  DFC  and  FBG are congruent in accordance with the  ASA-test of congruency of triangles.

It implies that the segments  DF  and  GF  are congruent as the corresponding sides of the congruent triangles  DFC  and  FBG.

Thus the mid-line  EF  of the trapezoid  ABCD  is the straight line segment connecting the midpoints of the triangle  AGD.

It is well known fact that the the straight line segment connecting the midpoints of the triangle  AGD  is parallel to the triangle base  AG  and its length is half of the length of the triangle base.  The line segment joining the midpoints of two sides of a triangle.

The length of the segment  EF  is half of the length  AG :   |EF| = 1/2*|AG| = 1/2F*(|AB| + |BG|).
Since  |BG| = |DC|  from the triangles congruency,  we have  |EF| = 1/2F2*(|AB| + |DC|),  or  |EF| = 1/2F2*(a + d),  where  a  and  d  are the lengths of the trapezoid bases.


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