Question

AB is a diameter of a circle with centre O and AC is it's chord Such that <BAC =30° , If the tangent drawn ath C, intersect extended angle at D, then show that BC=BD​

Answer 1

Answer:

∠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]