ABC is a right triangle at C,a perpendicular CD is drawn at AB prove AC^2/BC^2=AD/BD.?

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  • 1 decade ago
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    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

  • 4 years ago

    Find the distance between the points d, e, and f and if the distance is equal between all of them then they make an equilateral triangle.

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