Question

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in the adjoining figure, ABCD is a trapezium in which AB || DC.
If M and N are the mid points of AC and BD respectively, prove that MN =1/2(AB-CD).

[Hint: join CN and produce it to meet AB at E.
Then, △CDN is congruent to
△EBN. So,CD =EB and CN =NE .

MN,=1/2 AE (why?)=1/2 (AB-EB) =1/2(AB-CD)].


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Answer 1

Answer:

Given: ABCD is a trapezium in which AB || CD and M and N are mid-points of diagonal AC and BD respectively.

Construction: Join AN and produce it to meet CD at E.

To prove: MN =  1/2 (CD – AB) and MN || CD

Proof: In ∆ANB and ∆END

∠ANB = ∠END (vertically opposite angles)

NB = ND (N is the mid-points of BD)

and ∠ABN = ∠EDN (alternate angles)

(∵ AB || CD and BD is a transversal)

∆ANB = ∆END (by ASA congruency rule)

⇒ AN = NE and AB = ED …(i) (by c.p.c.t.c)

Now in ∆EAC,

N and M are the mid-points of AE and AC respectively.

MN || EC and MN =  1/2 EC (by mid-point theorem)

⇒ MN || EC

and MN =  1/2 (CD – ED) =  1/2 (CD – AB) [using (i)]

Hence, MN || CD (∵ MN || EC)

and MN =  1/2 (CD – AB)

Proved