Question

Using an appropriate diagram be able to prove the TriangleProportionality theorem and/or its converse.

Answer 1

The converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

From the sketch,

`(CD)/(DA)=(CE)/(EB)`

Take the reciprocal

`(DA)/(CD)=(EB)/(CE)`

`\text{Add 1 TO BOTH SIDES}`

`(DA)/(CD)+1=(EB)/(CE)+1`

Any number divided by itself is 1, so we can replace 1 with CD/CD or CE/CE

so that

`(CD)/(CD)+(DA)/(CD)=(CE)/(CE)+(EB)/(CE)`

Combine terms using our common denominator

`(CD+DA)/(CD)=(CE+EB)/(CE)`

from the diagram, we can see that

CA=CD+DA

and

CB=CE+EB

Then

`(CA)/(CD)=(CB)/(CE)`

Since the triangles have SAS for triangle similarity.

This means that

`\text{Triangle ABD is similar to Triangle }CDE`