Question

in the figure AB is a diameter and AC is chord​

Answer 1

Answer:

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∠ACB=90∘    [∠ from diameter]

In ΔACB

∠A+∠ACB+∠CBA=180∘

∠CBA=180∘−(90+30)

∠CBA=60∘ _________ (1)

In △OCB

OC=OB

so, ∠OCB=∠OBC         [opp sides are equal]

∴∠OCB=60∘

Now,

∠OCD=90∘

∠OCB+∠BCD=90∘

∠BCD=30∘ _______ (2)

∠CBO=∠BCO+∠CDB     [external ∠bisectors]

60=30+∠CDB

∠CDB=30∘ ________ (3)

from (2) & (3)

BC=BD     [ opp. ∠.S are equal]

Answer 2

Step-by-step explanation:

∠ACB=90∘ [∠ from diameter]

In ΔACB

∠A+∠ACB+∠CBA=180∘

∠CBA=180∘−(90+30)

∠CBA=60∘ _________ (1)

In △OCB

OC=OB

so, ∠OCB=∠OBC [opp sides are equal]

∴∠OCB=60∘

Now,

∠OCD=90∘

∠OCB+∠BCD=90∘

∠BCD=30∘ _______ (2)

∠CBO=∠BCO+∠CDB [external ∠bisectors]

60=30+∠CDB

∠CDB=30∘ ________ (3)

from (2) & (3)

BC=BD [ opp. ∠.S are equal]