AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects extended AB at a point D. Prove that BC = BD
Solution:
Given, AB is a diameter of a circle with centre O
AC is a chord of a circle.
Also, ∠BAC = 30°
The tangent at C intersects extended AB at a point D.
We have to prove that BC = BD.
We know that the angle between the tangent and the chord of a circle is equal to the angle made by the chord in the alternate segment.
So, ∠BCD = ∠BAC
Given, ∠BAC = 30°
Then, ∠BCD = 30° ------------------- (1)
We know that the angle in a semicircle is always equal to 90°
∠ACB = 90°
In triangle ABC,
∠CAB + ∠CBA + ∠ACB = 180°
30° + ∠CBA + 90° = 180°
120° + ∠CBA = 180°
∠CBA = 180° - 120°
∠CBA = 60°
From the figure,
Linear pair of angles,
∠CBA + ∠CBD = 180°
60° + ∠CBD = 180°
∠CBD = 180° - 60°
∠CBD = 120°
Considering triangle CBD,
We know that the sum of all three interior angles of a triangle is always equal to 180°
∠CBD + ∠BCD + ∠BDC = 180°
120° + 30° + ∠BDC = 180°
150° + ∠BDC = 180°
∠BDC = 180° - 150°
∠BDC = 30° ----------------------------- (2)
Comparing (1) and (2),
∠BCD = ∠BDC = 30°
We know that the sides opposite to equal angles are equal.
So, BC = BD
Therefore, it is proved that BC = BD
✦ Try This: Two chords AB and AC of a circle are equal. Prove that the centre of the circle lies on the bisector of angle BAC.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.4 Problem 8
AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects extended AB at a point D. Prove that BC = BD
Summary:
AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects extended AB at a point D. It is proven that BC = BD
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