In the given figure, ABCD is a trapezium in which AB || CD. M and N ar

1.	In the given figure, ABCD is a trapezium in which AB \|\| CD. M and N are mid-points of the diagonal AC and BD respectively. Prove that MN \|\| CD and MN = 1/2(CD – AB)
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 = \(\frac { 1 }{ 2 }\) (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) Now in ∆EAC, N and M are the mid-points of AE and AC respectively. MN \|\| EC and MN = \(\frac { 1 }{ 2 }\) EC (by mid-point theorem) ⇒ MN \|\| EC and MN = \(\frac { 1 }{ 2 }\) (CD – ED) = \(\frac { 1 }{ 2 }\) (CD – AB) [using (i)] Hence, MN \|\| CD (∵ MN \|\| EC) and MN = \(\frac { 1 }{ 2 }\) (CD – AB).

In the given figure, ABCD is a trapezium in which AB || CD. M and N are mid-points of the diagonal AC and BD respectively. Prove that MN || CD and MN = 1/2(CD – AB)

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 = \(\frac { 1 }{ 2 }\) (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)

Now in ∆EAC,

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

MN || EC and MN = \(\frac { 1 }{ 2 }\) EC (by mid-point theorem)

⇒ MN || EC

and MN = \(\frac { 1 }{ 2 }\) (CD – ED) = \(\frac { 1 }{ 2 }\) (CD – AB) [using (i)]