AB is a diameter of a circle_and AB poduced intersects at the point D the tangent to the circle at the point C Prove that angle BDC+ANGLE 2BCD=90degree
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Answer:
google pe search karo jaldi mil jayega solution
Answer:
REF.image
∠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]
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