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anonymous

  • one year ago

Hi I need help with this proof to prove Converse of the Side-Splitter Theorem: Given: XR over RQ = YS OVER SQ Prove: Line RS is parellel to Line XY Thanks

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  1. anonymous
    • one year ago
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    Here is a copy of the picture and the statements given.

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  2. anonymous
    • one year ago
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    For 5 the reason would be SAS similarity criterion

  3. anonymous
    • one year ago
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    Do you know what 6 & 7 would be?

  4. phi
    • one year ago
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    as nish states SAS (for similar triangles) see http://www.regentsprep.org/Regents/math/geometry/GP11/LsimilarProof.htm

  5. phi
    • one year ago
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    once you have shown two triangles are similar then you know "corresponding angles" are congruent.

  6. phi
    • one year ago
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    finally, if corresponding angles of a transversal are congruent, the lines are parallel

  7. anonymous
    • one year ago
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    Thanks

  8. anonymous
    • one year ago
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    Could you also help me with reason 2 & 3? I got those incorrect also. Thanks

  9. phi
    • one year ago
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    I would need a list of the reasons you have learned. in step 2, you add 1 to both sides (in the form RQ/RQ and SQ/SQ) the reason that is ok would read something like " equality remains true when you add equal amounts to both sides" for step 3, you replace XR+RQ with XQ (and also YS+SQ with YQ) the reason would be something like "the whole is the sum of its parts"

  10. mathstudent55
    • one year ago
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    reason 2 is not substitution

  11. anonymous
    • one year ago
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    Can you help with the Converse of the Side Splitter Theorem proof

  12. mathstudent55
    • one year ago
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    Reason 1. Given Reason 2. A property of proportions Reason 3. Segment addition postulate Reason 4. Congruence of angles is reflexive (this is a theorem) Reason 5. SAS Similarity Reason 6. Definition of similar triangles Reason 7. When two lines are cut by a transversal such that corresponding angles are congruent, then the lines are parallel. (this is a postulate)

  13. anonymous
    • one year ago
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    Thanks.

  14. mathstudent55
    • one year ago
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    You're welcome.

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