Question

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

Prove: CBF ≅ EDF using isometric (rigid) transformations.


Outline the necessary transformations to prove CBF ≅ EDF using a paragraph proof. Be sure to name specific sides or angles used in the transformation and any congruency statements.

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

Answer 2

The given triangle is rotated about 180 degrees. It can be said that the triangle CBF is congruent to the triangle EDF.

It is required to prove that triangle CBD is congruent to the triangle EDF using the isometric transformation.

Now, consider two parallel lines BC and DE. Point C is rotated by 180 degrees about the point F to make a point E. Similarly, rotate the point B by 180 degrees about the point F to make an image point D.

Now, the triangle EDF will be image of the triangle CBF because:

Point E is image of point C, Point D s image of point B and point F is common.

Therefore, it can be said that the triangle CBF is congruent to the triangle EDF.

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