centre of a circle is O and tangent line drawn from point P, PA and PB which touches the circle at A and B then prove that OP line segment is bisector of AB.
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In triangles PAC &PBC
PA=PB
angle APC=angle BPC
PC=PC
triangle PAC congruent triangle PBC [BY SAS]
=> AC=BC & angle ACP= angle BCP [CPCT]
bt, angle ACP+ angle BCP= 180 deg.
.`. angle ACP= angle BCP=90 deg.
Hence, OP is the perpendicular bisector of AB
Hope this helps.......
PA=PB
angle APC=angle BPC
PC=PC
triangle PAC congruent triangle PBC [BY SAS]
=> AC=BC & angle ACP= angle BCP [CPCT]
bt, angle ACP+ angle BCP= 180 deg.
.`. angle ACP= angle BCP=90 deg.
Hence, OP is the perpendicular bisector of AB
Hope this helps.......
muskan7171:
thanks..
please give you answer of my next question
from a exterior point P two tangent line PA and PB which touches the circle at A and B then prove that PAOB is a cyclic quadrilateral where O is the centre of circle.
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