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Circle theorem-Chord

  1. Nov 20, 2013 #1

    adjacent

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    One rule of circle theorems is,a line drawn from center to the mid-point of a chord cuts the cord at 90°.
    What's the proof?
    It's true that two radius and a chord creates an isosceles triangle.
    So, attachment.php?attachmentid=64135&stc=1&d=1384969206.gif

    How can I prove that in an isosceles triangle,a line drawn from the vertex angle to the mid-point of the base side cuts the side at 90°?
     

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    Last edited: Nov 20, 2013
  2. jcsd
  3. Nov 20, 2013 #2

    Nugatory

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    You don't (yet) know that those two angles are 90 degrees.... But what do you know about them?
     
  4. Nov 20, 2013 #3

    adjacent

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    I'm not talking about the base anlgles.I am talking about the 90° angle you see in the image.
    let vertex angle be y
    ##180-(\frac{y}{2}+x)=90##
    ##2x+y=180##

    I can't seem to solve y and x.I don't know whether this is a proof or not. :confused:
     
  5. Nov 20, 2013 #4

    Nugatory

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    I'm sorry, I wasn't clear. Those two 90-degree angles are the two angles that I mean, and my question still stands: What do you know about them?
     
  6. Nov 20, 2013 #5

    adjacent

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    hahahah.

    I don't know what do you mean.
    In a right angle triangle, ##c^2=a^2+b^2##

    c and b of both triangles is same.
    so if it is two right angle triangles,a should be same
    As the line was drawn to the midpoint,a is same in both triangles.So it is a right angle triangle.
    Is this enough for a proof?
     
    Last edited: Nov 20, 2013
  7. Nov 20, 2013 #6

    Nugatory

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    No, because you've worked in the assumption both are right triangles, which is what you're trying to prove. The ##c^2=a^2+b^2## relationship isn't helping any because you don't know the values for all three to prove that you do have a right triangle....

    But there's a reason I keep asking what you know about the two angles, not just one of them... What is their sum?
     
  8. Nov 20, 2013 #7

    adjacent

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    180°.Where any straight two lines intercept,any one side of the line has 180°.
    but why should this help?If we have a 40° and a 50° angle,we would still get 180 as a sum.
     
  9. Nov 20, 2013 #8

    HallsofIvy

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    Your midpoint line divides the large triangle into two smaller congruent triangles. They are congruent by "SSS": the two radii of the circle, the two bases, and the single ray in both triangles. From that follows that other "corresponding parts" are congruent.
     
  10. Nov 21, 2013 #9

    adjacent

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    I know it's congruent.but how do I know that the angle is 90°?
    I know that if vertex angle is y,the upper angle of both triangles have to be y/2
     
  11. Nov 21, 2013 #10
    Look at the attached figure.

    Since ΔABD is congruent to ΔACD ,

    ∠ADB = ∠ADC

    Now ∠ADB + ∠ADC =180° (Linear pair )

    So, 2∠ADB = 180° or ∠ADB = 90°
     

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  12. Nov 21, 2013 #11

    adjacent

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    Thank you.Now that makes sense.
    You all must be very skilled in proofs.I haven't even started.
     
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