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No-data

Let P a point outside the line XY, PO a perpendicular from P to XY, and PZ any line drawn from P to XY. Prove that PZ is not perpendicular to XY

  • 2 years ago
  • 2 years ago

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  1. No-data
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    |dw:1333151007374:dw|

    • 2 years ago
  2. No-data
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    |dw:1333151163362:dw|

    • 2 years ago
  3. AsianDuck
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    A triangle can not have more than one right angle.

    • 2 years ago
  4. No-data
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    Well I know that, but that sounds like a theorem and then I would have to prove it too. So double work =/

    • 2 years ago
  5. No-data
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    - Extend PO to P', so OP' = OP - Drawn P'Z - POP' is a line. - Then PZP' is not a line - then angle P'ZP does not have colinear sides.

    • 2 years ago
  6. No-data
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    On the other side - OP = OP' - OZ = OZ - triangle OPZ = triange OZP' - then angle OZP = angle OZP' - and angle OZP = (1/2)*angle P'ZP

    • 2 years ago
  7. No-data
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    I'm stuck there =/

    • 2 years ago
  8. dpaInc
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    you can try: PZO is a right triangle. and since POZ is a right triangle, OZP cannot be another right angle because then the sum of the two angles will be 180. and that leaves 0 degrees for the third angle.

    • 2 years ago
  9. No-data
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    That is a good Idea dpalnc but I haven't proof yet that the sum of the angles on an right triangle is 360º. Then I can't use it.

    • 2 years ago
  10. dpaInc
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    hmm... can you prove by contradiction?

    • 2 years ago
  11. No-data
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    Well I don't know how to prove by contradiction yet. I'm a newbie on the proving world =P

    • 2 years ago
  12. No-data
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    I don't think I should prove that kind of things, but the goal(of the book i'm studying with) is to develop the geometry from just a bunch of postulates.

    • 2 years ago
  13. dpaInc
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    proof by contradiction is kinda like what I first suggested. you take something that is obviously false and assume it to be true. and by the end of the proof you end up with concluding something ridiculous. so the assumption must be false.

    • 2 years ago
  14. dpaInc
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    anyway, that's cool that you develop geometry from scratch.

    • 2 years ago
  15. dpaInc
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    do you have the parallel postulate then?

    • 2 years ago
  16. No-data
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    mmm I don't think so what is that postulate?

    • 2 years ago
  17. TuringTest
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    you meant the sum of the angles in a triangle is \(180^{\circ}\) you want a proof for that? because if you accept that we can prove your thingy by contradiction easily

    • 2 years ago
  18. No-data
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    I don't want a proof of that @TuringTest. The thing is that on the book I'm reading proves this in the way I wrote down. The problem is that I don't understand the proof. =(

    • 2 years ago
  19. No-data
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    In other words I don't get why it concludes that is true after that series of propositions that I know are true.

    • 2 years ago
  20. TuringTest
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    ok, so can we accept that a triangle must have 3 angles that add to 180 ?

    • 2 years ago
  21. No-data
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    Yes if I accept that the proof is very easy. But my goal by now, is not to prove it but understand the proof I just copy.

    • 2 years ago
  22. TuringTest
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    let us assume that PZ is perpendicular to XY then PZ and OP can never intersect this contradicts the assumption that we drew PZ by connecting XY to P hence PZ is not perpendicular to XY

    • 2 years ago
  23. TuringTest
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    proof 2: let us call the triangle formed by the three line segments OPZ we know that <OPZ is a right angle let us assume that <OZP is also a right angle that leaves the third angle of the triangle as zero, hence it is not a triangle: a contradiction therefor <OZP is not a right angle, and PZ is not perpendicular to XY

    • 2 years ago
  24. No-data
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    That's impressive @TurinTest!

    • 2 years ago
  25. TuringTest
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    thanks, but they are both two very basic proofs by contradiction a little practice is all that's required

    • 2 years ago
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