side splitting theorem q

[needzmathhelp]

The side splitting theorem tells us the segments on the sides of the triangle are proportionate. But what is the relationship between the side segments and the parallel segments? Or am I just crazy...

In other words, whats the relationship between DB/DA
and DE/AC?

 

[tkhunny]

Similar triangles are similar triangles.

DB/DA is an odd comparison. DA isn't the side of a triangle. A better question might be BD/BA

 

[lookagain]

The side splitting theorem tells us the segments on the sides of the triangle are proportionate. But what is the relationship between the side segments and the parallel segments? Or am I just crazy...

In other words, whats the relationship between DB/DA
and DE/AC?

needzmathhelp,

working on differently shaped triangles and different levels of parallel segments,
I have noticed (not necessarily a fact, that is), that if the ratio of
DB/DA = a/b, then the ratio of DE/AC = a/(a + b).

This alleged fact would need a demonstration/proof, if it is, in fact, the case.

 

[Deleted member 4993]

The side splitting theorem tells us the segments on the sides of the triangle are proportionate. But what is the relationship between the side segments and the parallel segments? Or am I just crazy...

In other words, whats the relationship between DB/DA
and DE/AC?

AB/BD = AC/DE

(AD+BD)/BD = AC/DE

AD/BD + 1 = AC/DE

 

[wjm11]

The side splitting theorem tells us the segments on the sides of the triangle are proportionate. But what is the relationship between the side segments and the parallel segments? Or am I just crazy...
In other words, whats the relationship between DB/DA
and DE/AC?

Hello, Needzmathhelp,

I believe there are TWO DIFFERENT properties of the figure you have provided that you may be confusing with each other.

If line DE is parallel to line AC, then similar triangles are created: ABC ~ DBE. Corresponding sides of similar triangles are proportionate, so we can make statements such as BD/DE = BA/AC. That is one possible relationship between sides and parallel segments. Another would be BD/BA = DE/AC.

However, the “side splitting theorem” is a bit different. It tells us that BD/DA = BE/EC.

I hope that helps.