Question

Given: ("CB" ) ∥ ("ED" ) ; ("CB" ) ≅ ("ED" )

Answer 1

Answer:

CB // ED and both of  BD and CE are transversals

The point E will be the image of point C by rotation 180° around point F

And The point D will be the image of point B by rotation 180° around point F

Rotation is a kind of transformation

So,  ΔEDF will be the image of  ΔCBF

So, ΔCBF ≅ ΔEDF

Another way:

So, ∠B = ∠D  Alternate angles are congruent

and ∠C = ∠E  Alternate angles are congruent

So, ΔCBF and ΔEDF  have the following

1)  ∠C = ∠E  ⇒⇒⇒  Alternate angles are congruent (proved)

2) CB ≅ ED  ⇒⇒⇒ Given

2) ∠B = ∠D  ⇒⇒⇒ Alternate angles are congruent (proved)

From 1, 2 and 3

So, ΔCBF ≅ ΔEDF By SAS postulate