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

Question

anonymous · Answer

I have no clue :(

mathstudent55 · Answer

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mathstudent55 · Answer

Given triangle ABC with segment DE parallel to side B.

mathstudent55 · Answer

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anonymous · Answer

I don't get why the numbers are there if we just need to prove that the two sides are proportional @mathstudent55

mathstudent55 · Answer

Using corresponding angles and the parallel lines, angles 1 and 2 are congruent.
Also, angles 3 and4 are congruent.
By AA Similarity, triangles ABC and ADE are similar.
That makes the lengths of corresponding sides proportional.

$\dfrac{AD}{AB} = \dfrac{AE}{AC} $

By the segment addition postulate, we can rewrite segments AB and AC as sums:

$\dfrac{AD}{AD + DB} = \dfrac{AE}{AE + EC} $

mathstudent55 · Answer

I numbered the angles to make it easier to mention them instead of using three-letter names. Angle 1 is easier to write than angle ADE.

anonymous · Answer

ohh ok I see

mathstudent55 · Answer

We can now take the reciprocal of both sides:

$\dfrac{AD + DB}{AD} = \dfrac{AE + EC}{AE} $

By a property of proportions, we can conclude:

$\dfrac{DB}{AD} = \dfrac{EC}{AE} $

mathstudent55 · Answer

That proves it.

anonymous · Answer

thank you so much .. understand it a lot better now :)

mathstudent55 · Answer

You are welcome.