O is the center of the circle . PA and PB are two tangents of a circle of a point P. prove (i) PAOB is a cyclic quadrilateral (ii) po is the bisector of
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PAO = 90° [Tangent and the radius at the point of contact are perpenducular.]
Similarly, <PBO = 90° [Tangent and the radius at the point of contact are perpenducular.]
Consider the quadrilateral PAOB, <PAO + <PBO + <P + <O = 360° [Angle sum property.]
90°+ 90° + <P + <O = 360° <P + <O = 180°
Opposite angles of the quadrailteral PAOB are supplementary and hence it is a cyclic quadrailateral.
Join P and O.
Consider triangles PAO and PBO.
PA = PB [Tangents to a circle from a point are equal.]
PO = PO [Common side]
OA = OB [Radii]
So, triangles PAO and PBO are congrunet.
<APO = <BPO [CPCT]
⇒ PO is the bisector of the angle APB.
Similarly, <PBO = 90° [Tangent and the radius at the point of contact are perpenducular.]
Consider the quadrilateral PAOB, <PAO + <PBO + <P + <O = 360° [Angle sum property.]
90°+ 90° + <P + <O = 360° <P + <O = 180°
Opposite angles of the quadrailteral PAOB are supplementary and hence it is a cyclic quadrailateral.
Join P and O.
Consider triangles PAO and PBO.
PA = PB [Tangents to a circle from a point are equal.]
PO = PO [Common side]
OA = OB [Radii]
So, triangles PAO and PBO are congrunet.
<APO = <BPO [CPCT]
⇒ PO is the bisector of the angle APB.
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