In a figure, PT is a tangent to the circle and PBA is the secant .TM is the bisector pf angle ATB meeting AB at M .Prove that PT = PM
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As angle PTA = angle PTB ( tangent -angle theorem) (1)
And Angle BTM = angle MTA (given) (2)
And angle PMT = angle PBT + angle MTB
So angle PMT = angle PTA + angle MTA (using (1) and (2) )
And angle PTA + angle MTA = PTM
So angle PMT = angle PTM ( hence proved)
In triangle PMT
We have angle PMT = angle PTM
So PM = PT (isosceles triangle)
And Angle BTM = angle MTA (given) (2)
And angle PMT = angle PBT + angle MTB
So angle PMT = angle PTA + angle MTA (using (1) and (2) )
And angle PTA + angle MTA = PTM
So angle PMT = angle PTM ( hence proved)
In triangle PMT
We have angle PMT = angle PTM
So PM = PT (isosceles triangle)
Neha9088:
Thanks for the answer. But I didn't get how angle PTA = angle PTB. In the figure (which I sent along with the question) it seems that angle PTB is a part of the angle PTA
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