Question

"Question 11 In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see the given figure). Show that (i) Quadrilateral ABED is a parallelogram (ii) Quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) Quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF.

Class 9 - Math - Quadrilaterals Page 147" ​

Answer 1

Answer:

Refer the attachment:)

Answer 2

Answer:

(i)Suppose BD is a diagonal of quadrilateral ABCD.

As, AB=DE∴ ∠ABD=∠EDB(Alternate angles)

In ΔABD & ΔEDB

AB=ED(Given)

∠ABD=∠EDB

BD=BD(Common)

∴ΔABD≅ΔEBD (SAS rule)

Hence, BE=AD (CPCT)

Thus, in ABED Both pairs of opposite sides are equal

∴ABED is a parallelogram.

(ii)Again consider a diagonal BF of quadrilateral BCFE.

Similar to (i) we know that ΔBCE≅ΔEFC (SAS rule)

Hence, BE=CF (CPCT)

Thus, in BCFE Both pairs of opposite sides are equal

∴BCFE is a parallelogram.

(iii) From part(i), we proved that ABED is a parallelogram

So, AD=CF and AD∣∣CF

From part(ii), we proved that BEFC is a parallelogram

So, BE=CF and BE∣∣CF

hence from (i) and (ii), AD∣∣CF AD=CF

(iv)Again consider a diagonal AF of quadrilateral ADFC.

Similar to (i) we know that ΔACD≅ΔDFC (SAS rule)

Hence, AC=DF (CPCT)

Thus, in ADFC Both pairs of opposite sides are equal

∴ADFC is a parallelogram.

As ADFC is a parallelogram ∴ opposite sides are parallel also.

(v)Proved in (iv)

(vi)By above all, we know that AB=DE,BC=EF,AC=DF