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  • 14-05-2020
  • Mathematics
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Given: JL-tangent, LE-secant

Prove: LU·LE = LV·LK

Given JLtangent LEsecantProve LULE LVLK class=

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sqdancefan
sqdancefan sqdancefan
  • 14-05-2020

Explanation:

By the properties of secants and tangents, ...

  LU·LE = LJ²

  LV·LK = LJ²

By the transitive property of equality, ...

  LU·LE = LJ² = LV·LK

  LU·LE = LV·LK

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