PROVE THAT THE STRAIGHT LINE JOINING THE MID-POINTS OF THE DIAGONALS OF A TRAPEZIUM IS PARALLEL TO THE PARALLEL SIDES AND IS EQUAL TO HALF OF THEIR DIFFERENCE

given: ABCD is a trapezium. with parallel sides AB and CD.

E and F are the mid-points of the diagonals BD and AC respectively.

TPT: (1). and (2) EF= 1/2(DC-AB)

construction: join AE and produce to meet DC at G.

proof:

in the triangle AEB and triangle GED,

∠AEB=∠GED [opposite angles]

∠ABE=∠EDG [alternate interior angles]

BE= ED [given]

therefore by ASA congruency triangles are congruent.

therefore by CPCT

 AB=DG..........(1) and 

AE=EG.......(2) 

now in the triangle AGC,

E is the mid-point of AG, and F is the mid-point of AC.

therefore by the mid-point thm. 

and EF=1/2GC ..........(3)

GC=DC-DG

GC=DC-AB  [using (1)]

therefore

and EF=1/2[DC-AB]

hope this helps you.

cheers!!