Mathematics
AB is the diameter and AC is a chord of a circle with center O such that angle BAC = 30°. The tangent to the circle at C intersects AB produced in D. Show that : BC = BD.
Answer
The circle with center O is shown in the figure below:
We know that,
The angle between a tangent and chord through the point of contact is equal to an angle in the alternate segment.
∠BCD = ∠BAC = 30°.
Since, angle subtended by a segment at the center is double the angle suspended at the circumference.
∠BOC = 2∠BAC = 2 × 30° = 60°.
From figure,
∠DOC = ∠BOC = 60°.
∠OCD = 90° [As tangent at any point and radius through the point are perpendicular.]
In △OCD,
⇒ ∠DOC + ∠OCD + ∠CDO = 180° [By angle sum property of triangle]
⇒ 60° + 90° + ∠CDO = 180°
⇒ ∠CDO = 180° - 150° = 30°.
From figure,
∠BDC = ∠CDO = 30°.
In △BCD,
∠BDC = ∠BCD [Both = 30°]
∴ BC = BD [As sides opposite to equal angles are equal.]
Hence, proved that BC = BD.
Related Questions
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In the given figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find AE.
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In the figure given below, O is the center of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x, y and z.
