O is the centre of a circle. PA and PB are tangents to the circle from a point P. prove that
a) PAOB is a cyclic quadrilateral
b) PO is the bisector of angel APB
c) angel OAB = angel OPA

Dear Student,

Please find below the solution to the asked query:
aPBO=PAO=90°  Because tangent make 90° with radiusSoPBO+PAO=180°....1And sum of quadrilateral  is 360°So PBO+PAO+BPO+AOB=360°180°+BPO+AOB=360°BPO+AOB=360°-180°=180°.....2Because In cyclic quadrilateral Sum of opposite angles are 180°So with 1 and 2 PAOB is cyclic .   Hence ProvedbIn triangle OPB  and triangle OPAPBO=PAO=90°PO is commonBO=AO  radiusSO OPBOPASo POB=POA  So PO si bisector of APB    Hence provedcIn triangle ABOBO=OA  radius SO OAB=OBA.....4    Angle opposite to the same side is sameNow  as we prove in a PBOA is cyclic quadrilateralSo APO=ABO....5 Because angle by tha same chord is Now with 4 and 5OAB=OPA    Hence Proved

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  • 4

OB = OA (radii)
PA = PB (tangents from an external point)
<PBO = < PAO = 90o ( tangents are perpendicular to radius through point of contact )

1) <PAO + <PBO = 180
   so, <AOB + < APB = 360 - 180 = 180
hence, PAOB is cyclic.

2) Using RHS, we prove that triangle PAO is congruent to triangle PBO
so by cpct, <APO = <BPO,
so OP is bisector of <APB

  • 2
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