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Geometric deduction problem ?

  1. Apr 27, 2006 #1
    Please help me to solve the following question :
    ABCD is a parallelogram .BA is procude to X and BA=AX . Prove that angle of XCB is 90 degree.
     
  2. jcsd
  3. Apr 27, 2006 #2

    Curious3141

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    In general, no help is given unless you show some work first. But in this case, the problem looks wrong - the proposition you're required to prove doesn't even seem to be true in general. Draw a diagram and show some work first, then we can help you.
     
  4. Apr 27, 2006 #3

    HallsofIvy

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    I assume that by "procude" you mean to extend the line BA out to X.

    You can't prove this- it isn't true.

    Easy counterexample: Let ABCD be a square (which is a kind parallelgram). Let s be the common length of the sides. Extending BA to X such that the length of AB is equal to the length of AX means that the length of BX is 2s. Drawing the line segement CX gives a right triangle with legs of length s and 2s. The measure of the angle XCB is given by
    tan(XCB)= 2s/s= 2. Since that is finite, XCB is not a right triangle. (Or, more simply, XCB is an angle in a right triangle with right angle at B. A triangle cannot have two right angles!)
     
  5. Apr 29, 2006 #4
    But I draw it and it show it really 90 degree .See the diagram below :
     

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    Last edited by a moderator: Apr 29, 2006
  6. Apr 29, 2006 #5

    HallsofIvy

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    That has nothing to do with the problem you initially stated. Yes, it is possible to draw a parallelogram such that the angle XCB is a right angle.
    However, the problem, as you initially stated it was
    And that, in general, is not true.
     
    Last edited: Apr 29, 2006
  7. Apr 29, 2006 #6

    Curious3141

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    I can't even see the diagram.
     
  8. Apr 30, 2006 #7

    HallsofIvy

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    You didn't miss anything. It was a parallelogram (a rhombus, actually) carefully drawn to give the special case where the angle IS a right angle.
     
  9. May 2, 2006 #8
    But can I prove ?
     
  10. May 2, 2006 #9

    Curious3141

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    There's nothing to prove if the proposition is false.
     
  11. May 2, 2006 #10
    What do you mean by the proposition is false since the diagram show that the angle is really 90 degree.
     
  12. May 2, 2006 #11

    Curious3141

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    Certainly if your parallelogram is a rhombus with internal angles of 60 and 120 degrees, angle XCB is 90 degrees. That's the diagram you drew,

    But it isn't true in general! Nowhere in the original question were limits placed on the dimensions or internal angles of the parallelogram. In all the infinite number of possible cases where the conditions of the question are met, most of them will have angle XCB NOT equal to 90 degrees.

    Do you see ?
     
  13. May 3, 2006 #12
    That means that the angle is 90 degree only when the internal angle is 60 or 120 degree right ?
     
  14. May 3, 2006 #13

    Curious3141

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    For a rhombus, it only works if you draw it exactly like you did. If ABC is the 120 degree angle instead it doesn't work.

    You don't necessarily need a rhombus. It works in some parallelograms too.

    In parallelogram ABCD, the length BA = CD = l (length) and AD = BC = w (width) (as you drew it, except in your diagram, you drew a rhombus where l = w).

    Angle BCX is only 90 degrees if angle ABC = theta satisfies [tex]\cos\theta = \frac{w}{2l}[/tex]
     
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