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singhmmm

  • one year ago

A , B , E ARE POINTS ON CIRCLE AB IS DIAMETER WITH O AS A CENTRE AC AND BD ARE PERPENDICULARS ON A LINE PQ . BD MEETS THE CIRCLE AT E . PROVE THAT AC = ED

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  1. hba
    • one year ago
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    @singhmmm A diag would be actually helpful use the draw tool :) |dw:1357992984759:dw|

  2. mathslover
    • one year ago
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    well I think there are many possible diagrams. Since PQ line is not specified whether it lies inside the circle or outside or...

  3. hba
    • one year ago
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    The 9 point diag from the Halls book may resemble :P

  4. hba
    • one year ago
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    @singhmmm Would you mind replying ?

  5. mathslover
    • one year ago
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    well this is not an easy stuff.

  6. singhmmm
    • one year ago
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    @hba ..i didn' understand vat u r trying to say

  7. mathslover
    • one year ago
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    @singhmmm did you try this question ?

  8. singhmmm
    • one year ago
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    yup .

  9. singhmmm
    • one year ago
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    diagram is not making clear..

  10. singhmmm
    • one year ago
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    i don't know where to mark point E

  11. mathslover
    • one year ago
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    |dw:1357993427560:dw|

  12. mathslover
    • one year ago
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    since the question says that : i) BD meets the circle at E ii) PQ is a line ( nothing mentioned that it will be inside or outside the circle )

  13. mathslover
    • one year ago
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    r u sure that it is AC = ED ? @singhmmm

  14. singhmmm
    • one year ago
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    yup

  15. singhmmm
    • one year ago
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    somehow i think ques.statement is wrong

  16. singhmmm
    • one year ago
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    vat do u think

  17. mathslover
    • one year ago
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    it 'might' be but m confused something about the diagram. Not sure for wrong but I think something useful is missing

  18. singhmmm
    • one year ago
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    kk

  19. singhmmm
    • one year ago
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    thanks buddy

  20. mathslover
    • one year ago
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    Your welcome but I have invited some experts who can help us. Don't close this quest. now and stay updated

  21. mathslover
    • one year ago
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    I got it

  22. experimentX
    • one year ago
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    |dw:1357994247928:dw|

  23. mathslover
    • one year ago
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    Wait the diagram is wrong. Let me draw a correct one

  24. mathslover
    • one year ago
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    |dw:1357994306561:dw|

  25. mathslover
    • one year ago
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    |dw:1357994444885:dw| now let us join AE

  26. experimentX
    • one year ago
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    |dw:1357994482792:dw|

  27. mathslover
    • one year ago
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    by property that angle AEB will be 90 degree similarly all the angles in the quad. AEDC will be 90 degrees

  28. experimentX
    • one year ago
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    yep!! correct mathlover!!

  29. mathslover
    • one year ago
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    Therefore AEDC is a rectangle and AC = ED ( opposite sides of rectangle are equal )

  30. mathslover
    • one year ago
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    Thanks @experimentX

  31. singhmmm
    • one year ago
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    thank u so much @mathlover and experiment X

  32. mathslover
    • one year ago
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    you are welcome @singhmmm , hope you enjoy this site. Best of luck.

  33. singhmmm
    • one year ago
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    yup

  34. singhmmm
    • one year ago
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    sme 2 u

  35. mathslover
    • one year ago
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    thnks

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