Question

Please help. I will give a medal~



Prove that a line that divides two sides of a triangle proportionally is parallel to the third side. Be sure to create and name the appropriate geometric figures.

Answer 1

Given: \(\triangle ABC\)

\(~~~~~~~~~~\dfrac{AD}{DB} = \dfrac{AE}{EC}\)

Prove: \(\overline{DE} \parallel \overline{BC}\)

Answer 2

This is the proof you need to do.

Answer 3

Okay, but I don't know how to do it. Can you please show me how?

Answer 4

You need a 2-column proof?

Answer 5

ummm.. i don't think so? i think i just have to prove it.

Answer 6

Ok, then I will give you the outline of the proof.

We are given \(\dfrac{AD}{DB} = \dfrac{AE}{EC} \)

Through a property of proportions, we can say \(\dfrac{AD}{AD + DB} = \dfrac{AE}{AE + EC} \)

By the segment addition postulate,
AD + DB = AB, and AE + EC = AC

Answer 7

If you are following what I am writing, notice that I just corrected what I wrote just above.

Answer 8

By substitution,

\(\dfrac{AD}{AB} = \dfrac{AE}{AC} \)

The proportion above shows that the lengths of the sides of triangles ADE and ABC are proportional. Also, we know that angle A is congruent to angle A.
Then by the SAS Similarity theorem, triangles ABC and ADE are similar.

Answer 9

Okay, that makes sense.

Answer 10

Since those triangles are similar, then by the definition of similar triangles, angles ADE and ABC are congruent.

Finally, lines DE and BC are parallel because of the postulate: "If two lines are cut by a transversal such that corresponding angles are congruent, then the lines are parallel."

Answer 11

sorry, but gtg

Answer 12

thank you! this helped a lot! :)

Answer 13

you're welcome