ABC is a triangle with angle C =90*.CD is drawn &D lies on AB such that angleCDA=angleCDB=90*.?

Prove that AC^2/BC^2=AD/BD

Update:

no proper reply yet. where are the top contributers?

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  • 1 decade ago
    Favourite answer

    I think you have asked this question twice i have already answered this anyhow here is the proof

    in triangle ACD

    angle ACD + angle CAD = 90

    in triangle ACB

    angle CAB + ABC = 90

    angle CAB = angle CAD same angle

    therefore angle ACD = angle ABC --------------------------------(1)

    consider the Triangles

    ACD and CBD

    angle ACD = angle CBD | angle ABC = angle CBD

    angle ADC = angle CDB

    so the two triangles are similar

    so we have

    AD/CD = AC/CB = DC/DB -----------------------------------2)

    AD/DC = AC/BC

    DC/BD = AC/BC | from 2

    multiplying both

    AD/DC * DC/BD = AC/BC * AC/BC

    AD /BD = AC^2/BC^2 prooved

  • 1 decade ago

    we know that in triangle ABC AC^2/BC^2=DC^2+AD^2/CD^2+BD^2

    =>YOUR ANSWER

    Source(s): general
  • Anonymous
    1 decade ago

    Math is dull lol

  • 1 decade ago

    right

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