Question

Two tangents are drawn to a circle with centre O from an external point A. Let the points of contact be B and C respectively. BO is joined and produced to meet the circumference at D. A perpendicular is drawn from C on BD at E. Prove that AD when joined bisects CE.​

Answer 1

Answer:

We know that, tangent perpendicular to the radius.

∴  ∠BCO=∠BDO=90o            --- ( 1 )

In △OBC and △OBD

⇒  ∠BCD=∠BDO=90o           [ from ( 1 ) ]

⇒  OB=OB                [ Common side ]

⇒  OC=OD               [ Radius of a circle ]

∴  △OBC≅△OBD         [ By R.H.S congruence rule ]

⇒  ∠OBC=∠OBD        [ C.P.C.T ]

∴  ∠OBC=∠OBD=60o 

In △OBC,

⇒  cos60o=OBBC

⇒  21= $\frac{OB}{BC}$

⇒  OB=2BC