In Δs ABC and DEF, AB||DE; BC=EF and BC||EF. Vertices A, B and C are

1.	In Δs ABC and DEF, AB\|\|DE; BC=EF and BC\|\|EF. Vertices A, B and C are joinedto vertices D, E and F respectively (see figure). Show that(i) ABED is a parallelogram(ii) BCFE is a parallelogram(iii) AC=DF(iv) ΔABC≅ΔDEF
Answer» Hey there! -------- *(i)* *ABED is a PARALLELOGRAM.* Given : △ABC and △DEF, AB = DE, AB II DE, BC = EF and BC = EF To Prove : ABED is a parallelogram. Proof : Given that AB = DE and AB II DE => [One pair of opposite sides is PARALLEL and equal.] Hence, ABED is a parallelogram. *(ii)* *BCFE is a parallelogram.* Proof : Given that BC = EF and BC II EF => [One pair of opposites sides is parallel and equal.] Hence, BCFE is a parallelogram. *(iii)* *AC = DF* Proof : From (i) and (ii) we get, AD = CF and AD II CF => [One pair of opposite sides are equal and parallel to each other.] Therefore, ACFD is a parallelogram. So, AC = DF ...........[Opposite sides of parallelogram] *(IV)* *△ABC ≅ △DEF* Proof : AB = DE ........(given) BC = EF ........(given) AC = DF .......[proved in (iii)} Hence, △ABC ≅ △DEF ..............[By SSS congruence rule]

In Δs ABC and DEF, AB||DE; BC=EF and BC||EF. Vertices A, B and C are joinedto vertices D, E and F respectively (see figure). Show that(i) ABED is a parallelogram(ii) BCFE is a parallelogram(iii) AC=DF(iv) ΔABC≅ΔDEF

Answer»

Hey there!

--------

(i) ABED is a PARALLELOGRAM.

Given :

△ABC and △DEF,

AB = DE, AB II DE, BC = EF and BC = EF

To Prove : ABED is a parallelogram.

Proof :

Given that AB = DE and AB II DE

=> [One pair of opposite sides is PARALLEL and equal.]

Hence, ABED is a parallelogram.

(ii) BCFE is a parallelogram.

Proof :

Given that BC = EF and BC II EF

=> [One pair of opposites sides is parallel and equal.]

Hence, BCFE is a parallelogram.

(iii) AC = DF

Proof :

From (i) and (ii) we get,

AD = CF and AD II CF

=> [One pair of opposite sides are equal and parallel to each other.]

Therefore, ACFD is a parallelogram.

So, AC = DF ...........[Opposite sides of parallelogram]

(IV) △ABC ≅ △DEF

Proof :

AB = DE ........(given)

BC = EF ........(given)

AC = DF .......[proved in (iii)}

Hence, △ABC ≅ △DEF ..............[By SSS congruence rule]