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mahmit2012

Q10: What's wrong in this question?

  • one year ago
  • one year ago

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  1. mahmit2012
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    |dw:1340318787218:dw|

    • one year ago
  2. mahmit2012
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    suppose to: ABCD is rectangular AE=AD PK and PH are perpendicular bisector EK=KC , DH=HC In conclusion: AP=BP AE=BC PE=PC <PAB=<PBA Then <PAE=<PBC Or <EAB=<CBA=90...!!!!!

    • one year ago
  3. maheshmeghwal9
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    Omg mahmit sir O.O

    • one year ago
  4. mahmit2012
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    Hi , can you solve the problem above?

    • one year ago
  5. mahmit2012
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    It shows 120=90 !!

    • one year ago
  6. ganeshie8
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    if AD = AE, PE has to be more than PC. both cannot be same.. so question is wrong.

    • one year ago
  7. ganeshie8
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    u see ? think of a circle with radius AD and center at A.. since P is fixed point, its length changes as it moves along the circle..

    • one year ago
  8. mahmit2012
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    PK is perpendicular bisector and shows PE=PC what's wrong?

    • one year ago
  9. mahmit2012
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    no P change with E it obviously depend to E

    • one year ago
  10. ganeshie8
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    wait.. im confused again.... gimme some time il get back

    • one year ago
  11. mahmit2012
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    ok thanks

    • one year ago
  12. ganeshie8
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    PD = PC (P is on perpendicular bisector of DC)

    • one year ago
  13. mahmit2012
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    yes it is

    • one year ago
  14. ganeshie8
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    i see it some what blurred way.. let me explain my logic

    • one year ago
  15. ganeshie8
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    with P as center, and PD as radius draw a circle. it goes through D and C points

    • one year ago
  16. ganeshie8
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    draw another circle, with A as center, and AD as radius

    • one year ago
  17. ganeshie8
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    |dw:1340380572369:dw|

    • one year ago
  18. mahmit2012
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    Do you think points P A and E are in a straight line?

    • one year ago
  19. ganeshie8
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    no way lol... PAE & PBC are congruent triangles i can see that

    • one year ago
  20. ganeshie8
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    the thing im trying to figure out and put is : two circles intersect at two no more than two points...

    • one year ago
  21. ganeshie8
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    |dw:1340381146334:dw|

    • one year ago
  22. ganeshie8
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    with respect to PA straight line, if P is fixed, and PE = PD, PE and PD lie on either side of PA

    • one year ago
  23. ganeshie8
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    im not sure.. if this is satisfying for u. but its ok to me to conclude that question is wrong based this....

    • one year ago
  24. ganeshie8
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    |dw:1340383701791:dw|

    • one year ago
  25. ganeshie8
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    so if we move E to the other side of PA straight line, triangles EAP, DAP, CBP are congruent

    • one year ago
  26. mahmit2012
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    when you move E should change EC perpendicular bisector then K and P will be different.

    • one year ago
  27. ganeshie8
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    ok did i make myself clear why i think PE should come other side of PA ? in your diagram both PE and PD are on same side of PA which is wrong

    • one year ago
  28. mahmit2012
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    If you have cad to draw exactly the figure please try to figure out what happen in there? Thank you for helping and hope we will get a logical proofing after all.

    • one year ago
  29. ganeshie8
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    |dw:1340392907050:dw|

    • one year ago
  30. mahmit2012
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    |dw:1340392825081:dw|

    • one year ago