Let A,B,T be three distinct points on a circle, and P be a point on the extension of BA. Prove that PT is a tangent to the circle iff $PT^2=PA.PB$. I have proved the part "if PT is tangent then $PT^2=PA.PB$", but, yet can't prove "if $PT^2=PA.PA$ then PT is tangent"
please, give a hints. If i failed, then give the solution!
Tangent
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Re: Tangent
Oh, i finally solved it. But, won't leaving any solution for trying urself!
A man is not finished when he's defeated, he's finished when he quits.
Re: Tangent
I think that now you can post the solution.
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Re: Tangent
i have proved the main and inverse like Hasib and giving both.
main part:
PT either is a tangent or crosses the circle.suppose PT intersects the circle at M different from P.
so,
\[PT.PM=PA.PB\]
but,
\[PT^2=PA.PB\]
so,
\[PT=PM\]
it's not possible.that means PT doesn't intersect circle.it is tangent.
inverse part:
add A,P and B,P.in triangle ATP and PTB
angle APT is common.
\[\angle ATP=\angle TBP\]
EKANTOR BRITTANGSHOSTO ANGLE
so these triangles are similar.make ratio of the sides and you will get easily $PT^2=PA.PB$
main part:
PT either is a tangent or crosses the circle.suppose PT intersects the circle at M different from P.
so,
\[PT.PM=PA.PB\]
but,
\[PT^2=PA.PB\]
so,
\[PT=PM\]
it's not possible.that means PT doesn't intersect circle.it is tangent.
inverse part:
add A,P and B,P.in triangle ATP and PTB
angle APT is common.
\[\angle ATP=\angle TBP\]
EKANTOR BRITTANGSHOSTO ANGLE
so these triangles are similar.make ratio of the sides and you will get easily $PT^2=PA.PB$
Try not to become a man of success but rather to become a man of value.-Albert Einstein
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Re: Tangent
i think i have forgotten everything of school geometry.can you please explain.
Re: Tangent
Hi, KHADIJA
it would be better if u glance over the geometry books of 9-10 and read the theorems carefully......
it would be better if u glance over the geometry books of 9-10 and read the theorems carefully......
A man is not finished when he's defeated, he's finished when he quits.
Re: Tangent
This is problem 9 in exercise 4 of Secondary Higher geometry book