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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 .

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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 .

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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 .

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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 .

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const. - extend CD to point P where it will meet a line drawn from point B passing through point E.

proof- in triangles PED ABE,

angle DEP=angle AEB (vert. opp. angles)

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

angle EDP=angle BAE (alt. angles).

therefore,by ASA congruency , triangle PED is congruent to triangle AEB .

HENCE , AB = DP(CPCT)

BE = EP(CPCT)

therefore E is the midpoint of BP.

NOW,

EF are mid-points of PBBC respectively.

therefore EF II CP EF=1/2 CP (by mid-point theorem).

EF = 1/2 CP

EF=1/2 (CD+DP)

EF=1/2 (CD+AB)

Hence , proved .

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ABCD is a trapezium in which AB || DC and E, F are mid points of AD, BC respectively.

Join CE and produce it to meet BA produced at G.

In ΔEDC and ΔEAG,

ED = EA    ( E is mid point of AD)

∠CED = ∠GEC ( Vertically opposite angles)

∠ECD = ∠EGA ( alternate angles) ( DC||AB, DC||GB and CG transversal)

∴ ΔEDC ≅ ΔEAG

CD  = GA and EC = EG 

In ΔCGB,

E is mid point of CG ( EC = EG proved)

F is a mid point of BC  (given)

∴ By mid point theorem EF ||AB and EF = (1/2)GB.

But GB = GA + AB = CD + AB

Hence EF||AB and EF = (1/2)( AB + CD).
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e and f are respectively the midpoints of non-parllel sides of ad and bc of a trapezium abcd prove that  ef parllel to ab
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