Question

answer the question 30

Answer 1

Given =>Since, the tangent at any point of a circle
is perpendicular to the radius through the point
of contact.
2OAP LOBP = 90°.(i)
In right AOAP and AOBP,
OP = OP (common)
OA OB (radii of circle)
LOAP = ZOBP = 90° [from (i)]
:: A 0AP - A OBP
(Using R.H.S. congruent condition)
=AOP => BOP
Now, in triangles OAM and OBM,
OA OB (radii of circle)
ZAOM = LBOM [ ZAOM ZAOP and 2BOM
BOP]
and OM OM (common)
AOBM AOBM [Using SAS congruent condition]
AM BM.(i)
and 2QMA 2OMB
But OMA 2OMB 180° (linear pair)
24OMA 180° ZOMA 90°
OM I AB ...(ii)
From (i) and (ii)
OP is the perpendicular bisector of AB.