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AB is a diameter and AC is a chord of a circle with centre O such that angle BAC=30 the tangent to the circle at C intersectsAB produced in D show that BC=BD

Posted by John Stone 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Given: A circle with centre O. A tangent CD at C.
Diameter AB is produced to D.
BC and AC chords are joined, ∠BAC = 30°

To prove: BC = BD
Proof: DC is tangent at C and, CB is chord at C.
Therefore, ∠DCB = ∠BAC [∠s in alternate segment of a circle]
⇒ ∠DCB = 30° …(i) [∵ ∠BAC = 30° (Given)]
AOB is diameter. [Given]
Therefore, ∠BCA = 90° [Angle in s semi circle]
Therefore, ∠ABC = 180° - 90° - 30° = 60°
In ΔBDC,
Exterior ∠B = ∠D + ∠BCD
⇒ 60° = ∠D + 30°
⇒ ∠D = 30° …(ii)
Therefore, ∠DCB = ∠D = 30° [From (i), (ii)]
⇒ BD = BC [∵ Sides opposite to equal angles are equal in a triangle]
Hence, proved.

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