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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    I have no clue :(

  2. mathstudent55
    • one year ago
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    |dw:1437585347423:dw|

  3. mathstudent55
    • one year ago
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    Given triangle ABC with segment DE parallel to side B.

  4. mathstudent55
    • one year ago
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    |dw:1437585463132:dw|

  5. anonymous
    • one year ago
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    I don't get why the numbers are there if we just need to prove that the two sides are proportional @mathstudent55

  6. mathstudent55
    • one year ago
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    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: \(\dfrac{AD}{AD + DB} = \dfrac{AE}{AE + EC} \)

  7. mathstudent55
    • one year ago
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    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.

  8. anonymous
    • one year ago
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    ohh ok I see

  9. mathstudent55
    • one year ago
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    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: \(\dfrac{DB}{AD} = \dfrac{EC}{AE} \)

  10. mathstudent55
    • one year ago
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    That proves it.

  11. anonymous
    • one year ago
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    thank you so much .. understand it a lot better now :)

  12. mathstudent55
    • one year ago
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    You are welcome.

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