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


Hope this information will clear your doubts about the topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards

  • 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
What are you looking for?