Question

please proof theorem 10.1

Answer 1

$\underline\bold{\huge{ANSWER \: :}}$

▶GIVEN : A Circle was drawn and a tangent AB was also drawn meeting with the circle at point P with the centre as O.

▶TO PROVE : OP is perpendicular to AB.

▶ CONSTRUCTION : Take any point Q, other than P, on the tangent AB. Join O, Q. Suppose OQ meets the circle at R.

▶ PROOF :

OP = OR (Radii of same circle)
OQ = OR + RQ

=> OQ > OR

=> OQ > OP [SINCE, OP = OR]

=> OP < OQ

Thus, OP is shorter than any other segment joining O to any point of AB. So, this is the least distance and least distance is the perpendicular distance.

Hence, OP is perpendicular to AB.

Answer 2

given= a circle c(o.r)meet tangent AB at P
construct: take a pt Q on tangent AB other than P
to prove:op perpendicular AB
proof:OP=OR(radii of circle)
OQ=OR+RQ
OQ>OR
OQ>OP
OP<OQ
there for Op is the shortest side
as perpendicular is the shortest side
there for OP is perpendicular to AB