Question

if EF II AB . then prove that EF =1/2 (AB + DC) when E and F are the mid points of non parallel side AD and BC .

Answer 1

Step-by-step explanation:

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Answer 2

ANSWER:-

Given:

ABCD is a trapezium in which EF||AB & E ,F are mid-point of AD & BC respectively.

To prove:

EF= 1/2(AB+DC).

Construction:

Join CE and extend to G which is produced from BA.

Proof:

In ∆EDC & ∆EAG,

ED = EA [E is the mid-point of AD]

∠CED = ∠GEC [vertically opposite angle]

∠ECD = ∠EGA [alternate angles]

[DC||AB, DC||GB & CG is transversal]

Therefore,

∆EDC ≅ ∆EAG [ASA congruence rule]

CD = GA & EC= EG

In ∆CGB,

E is the mid-point of CG [EC=EG proved]

F is the mid-point of BC [given]

Therefore,

By midpoint theorem EF||AB & EF=1/2GB

But GB= GA+AB= CD+AB

Hence,

EF||AB

EF= 1/2(AB+DC).

Proved.

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