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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

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Please find below the solution to the asked query:
a∠PBO=∠PAO=90°  Because tangent make 90° with radiusSo∠PBO+∠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 OPA⇒∠PBO=∠PAO=90°⇒PO is common⇒BO=AO  radiusSO △OPB≅△OPASo ∠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 5⇒∠OAB=∠OPA    Hence Proved


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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

 
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