anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Please explain with diagram.
nenadmatematika
  • nenadmatematika
go angela go :D
anonymous
  • anonymous
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.

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anonymous
  • anonymous
I don't think it will help in getting the solution.
anonymous
  • anonymous
Okay I got it but my mouse is not working properly.. If somebody could redraw a good figure labelled??
anonymous
  • anonymous
Wait. I am drawing the figure.
anonymous
  • anonymous
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anonymous
  • anonymous
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.
anonymous
  • anonymous
Anyone der? Please reply soon.
anonymous
  • anonymous
so you need a theorem proof?
anonymous
  • anonymous
do you know any of the angles?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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?
Mani_Jha
  • Mani_Jha
Take the point of intersection of AB and OP as I. Can you prove that OI=IQ=r/2, Aadarsh?
anonymous
  • anonymous
But how?
Mani_Jha
  • Mani_Jha
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?
Mani_Jha
  • Mani_Jha
Hey pls tell me if you got it
anonymous
  • anonymous
Can u show the steps plzzzzzzzzzzzzz?
kunal
  • kunal
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
Mani_Jha
  • Mani_Jha
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.
anonymous
  • anonymous
Thanks Kunal and Mani bhaiiiiiiii. Short and sweet proof. Cheers Kunal!!!!!!!!!

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