Question

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Answer 1

Answer:

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Step-by-step explanation:

Given : OO is the centre of the circle.

Seg ABAB is a diameter, CDCD is the tangent at CC.

BDBD is a tangent at BB

In ΔCODΔCOD and ΔBODΔBOD

⇒OC=OB⇒OC=OB    ....(Radii of same circle)

⇒CD=DB⇒CD=DB   ....(Tangent from an external point)

⇒OD=OD⇒OD=OD    ....(Common side)

⇒ΔCOD=ΔBOD⇒ΔCOD=ΔBOD      (SSS test)

⇒ΔCOD=ΔBOD=x⇒ΔCOD=ΔBOD=x     (c.a.s.t.) equation (i)

In ΔAOCΔAOC,  AO=OCAO=OC   ....(Radii of same circle)

⇒⇒  ∠OAC=∠OCA=y∠OAC=∠OCA=y  ....(Isoceles ΔΔ theorem) equation (ii)

⇒∠COB=∠OAC+∠OCA⇒∠COB=∠OAC+∠OCA   ....(Remote interior ∠∠ theorem)

⇒∠COD+∠BOD=∠OAC+∠OCA⇒∠COD+∠BOD=∠OAC+∠OCA

⇒x+x=y+y⇒x+x=y+y   ....[From equation (i)  and (ii)]

⇒2x=2y⇒2x=2y

⇒x=y⇒x=y   ....(dividing by 22)

⇒∠COD=∠OCA⇒∠COD=∠OCA

⇒⇒  Seg OD∥OD∥ chord ACAC.   ....(Alternate ∠s∠s test)             

Hence, proved.