Question

in the adjoining figure, O is the centre and seg AB is a diameter. At the point C on the circle, the tangent CD is drawn. Line BD is a tangent to the circle at the point B. Show that seg OD ||chord AC​

Answer 1

Step-by-step explanation:

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Answer 2

seg OD || chord AC​

Step-by-step explanation:

in Δ OCD  &  Δ OBD

OC = OB  ( radius)

CD = BD ( common tangent)

OD = OD common  

=>  Δ OCD  ≅  Δ OBD

=> ∠COD = ∠BOD

in ΔAOC

OA = OC  ( radius)

=> ∠OCA = ∠OAC

∠COB = ∠OCA +  ∠OAC ( extriore angle of ΔAOC)

=> ∠COD + ∠BOD = ∠OCA  + ∠OCA

=> ∠COD + ∠COD = 2∠OCA

=> 2∠COD = 2∠OCA

=> ∠COD = ∠OCA

=> AC ║ OD  ( alternate angle)

QED

Proved

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