Question

In the adjoining figure AB is in diameter of
circle with centre O and a circle is described
with Ao as diameter. A chord AD of the bigger
circle intersects the smaller circle at C. Prove
that BD = 2 xoc.


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Answer 1

Answer:

In △BDA and △OCA

∠BAD=∠OAD=45

o

(common)

∠BDA=∠OCA=90

o

(angle in semicircle)

Hence,

△BDA and △OCA are similar (by AA)

OA

BA

=

OC

BD

=

AC

AD

and BA=2r,OA=r

OC

BD

=

r

2r

⇒BD=2OC