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Aadarsh
 2 years ago
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.
Aadarsh
 2 years ago
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.

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Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0Please explain with diagram.

nenadmatematika
 2 years ago
Best ResponseYou've already chosen the best response.0go angela go :D

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0Angela sister, ur figure is somewhat correct. But, we should first join OA and OB separately. Then, join AB as a separate linesegment for the triangle PAB.

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0I don't think it will help in getting the solution.

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0Okay I got it but my mouse is not working properly.. If somebody could redraw a good figure labelled??

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0Wait. I am drawing the figure.

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0The 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.

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0Anyone der? Please reply soon.

Sandeez
 2 years ago
Best ResponseYou've already chosen the best response.0so you need a theorem proof?

Sandeez
 2 years ago
Best ResponseYou've already chosen the best response.0do you know any of the angles?

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0No 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.

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0AQ 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
 2 years ago
Best ResponseYou've already chosen the best response.1Take the point of intersection of AB and OP as I. Can you prove that OI=IQ=r/2, Aadarsh?

Mani_Jha
 2 years ago
Best ResponseYou've already chosen the best response.1Take OI=x. IP is this, (2rx). 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
 2 years ago
Best ResponseYou've already chosen the best response.1Hey pls tell me if you got it

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0Can u show the steps plzzzzzzzzzzzzz?

kunal
 2 years ago
Best ResponseYou've already chosen the best response.2its 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
 2 years ago
Best ResponseYou've already chosen the best response.1Will you accept that AP=r*sqrt3? \[AI ^{2}=PA ^{2}PI ^{2}\] =\[3r ^{2}(2rx)^{2}\] \[AI ^{2}=AO ^{2}OI ^{2}\] =\[r^{2}x ^{2}\] \[3r ^{2}(2rx)^{2}=r ^{2}x ^{2}\] Now solve this, u get x=r/2.

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks Kunal and Mani bhaiiiiiiii. Short and sweet proof. Cheers Kunal!!!!!!!!!
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