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Aadarsh Group Title

From a point "P", 2 tangents, PA and PB are drawn to a circle with centre "O". If OP is equal to diameter of circle, prove that triangle PAB is an equilateral triangle.

  • 2 years ago
  • 2 years ago

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  1. Aadarsh Group Title
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    Please explain with diagram.

    • 2 years ago
  2. nenadmatematika Group Title
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    go angela go :D

    • 2 years ago
  3. Aadarsh Group Title
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    Angela sister, ur figure is somewhat correct. But, we should first join OA and OB separately. Then, join AB as a separate line-segment for the triangle PAB.

    • 2 years ago
  4. Aadarsh Group Title
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    I don't think it will help in getting the solution.

    • 2 years ago
  5. siddhantsharan Group Title
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    Okay I got it but my mouse is not working properly.. If somebody could redraw a good figure labelled??

    • 2 years ago
  6. Aadarsh Group Title
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    Wait. I am drawing the figure.

    • 2 years ago
  7. Aadarsh Group Title
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    |dw:1330584831022:dw|

    • 2 years ago
  8. Aadarsh Group Title
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    The only thing I have to prove is : Prove that AQ = OQ = AO = r, where "r" is the radius. If u can prove AQ = radius, I can do the rest.

    • 2 years ago
  9. Aadarsh Group Title
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    Anyone der? Please reply soon.

    • 2 years ago
  10. Sandeez Group Title
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    so you need a theorem proof?

    • 2 years ago
  11. Sandeez Group Title
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    do you know any of the angles?

    • 2 years ago
  12. Aadarsh Group Title
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    No angle. But I know that angle OAP = 90 degrees. See, in triangle, OAQ, OA = OQ = radius So, it is an isosceles triangle. Now, if we prove that AQ = radius, then it becomes an equilateral triangle.

    • 2 years ago
  13. Aadarsh Group Title
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    AQ will be equal to radius, that's what our teacher told. She said, line segment from A on hypotenuse OP at its midpoint will be equal to the radius. But, what's the justification? What reason should I write?

    • 2 years ago
  14. Mani_Jha Group Title
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    Take the point of intersection of AB and OP as I. Can you prove that OI=IQ=r/2, Aadarsh?

    • 2 years ago
  15. Aadarsh Group Title
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    But how?

    • 2 years ago
  16. Mani_Jha Group Title
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    Take OI=x. IP is this, (2r-x). Using pythagoras' theorem, find AI from triangle AIP(AP=r*sqrt3) and then from triangle AOI. Equate those two. You will get x=r/2. After that, prove the congruency of AOI and AQI. You will get AQ=r. Do I need to show you the steps?

    • 2 years ago
  17. Mani_Jha Group Title
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    Hey pls tell me if you got it

    • 2 years ago
  18. Aadarsh Group Title
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    Can u show the steps plzzzzzzzzzzzzz?

    • 2 years ago
  19. kunal Group Title
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    its easy boss................... see..... we have PO=2OA => in triangle OAP, sin(angle APO) = OA/OP => sin(angel APO) = OA/2OA = 1/2 = sin 30 => angle APO = 30 also, angle APO= angle OPB, => angle APB = 60 also AP =PB => angle ABP = angle BAP by angle sum property, all angles are 60 then its a eqvilateral triangel

    • 2 years ago
  20. Mani_Jha Group Title
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    Will you accept that AP=r*sqrt3? \[AI ^{2}=PA ^{2}-PI ^{2}\] =\[3r ^{2}-(2r-x)^{2}\] \[AI ^{2}=AO ^{2}-OI ^{2}\] =\[r^{2}-x ^{2}\] \[3r ^{2}-(2r-x)^{2}=r ^{2}-x ^{2}\] Now solve this, u get x=r/2.

    • 2 years ago
  21. Aadarsh Group Title
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    Thanks Kunal and Mani bhaiiiiiiii. Short and sweet proof. Cheers Kunal!!!!!!!!!

    • 2 years ago
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