E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD.Prove that EF is parallel to AB and EF=1/2(AB+CD)

CONST. - Extend DC to point P where it meets a line drwn from point B through point E .

PROOF - In triangles DEP and AEB ,

AngleDEP = AngleAEB (vert.opp.angles)

AE=ED(E is the mid point of AD)

AngleBAE=AngleEDP(alt. angles)

Therefore, by ASA congruence , triangle ABE is congruent to triangle PED .

Hence AB=DP BE=EP(C.P.C.T.)

Therefore, E is the mid point of BP .

Now, in triangle BPC ,

E F are the mid points of PB BC respectively .

Therefore EF is parallel to CD EF = 1/2 CP

EF = 1/2 CP

EF=1/2 (CD + DP )

EF=1/2 (CD + AB) as AB =DP.

Hence , proved .