NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 9: Circles - Exercise 9.4Question 8
Class 10 - Mathematics - CHAPTER 9: Circles - Exercise 9.4
Question. 8
AB is a diameter and AC is a chord of a circle with centre O such that \(\angle BAC = 30^\circ\). The tangent at C intersects extended AB at a point D. Prove that \(BC = BD\).
Answer:
\(BC = BD\)
Detailed Answer with Explanation:
Step 1: Recall that when a line touches a circle at a point (here at C), the line is called a tangent.
Step 2: By the tangent–chord theorem, the angle between the tangent at C and the chord AC is equal to the angle in the opposite arc. That means: \(\angle ACD = \angle ABC\).
Step 3: Since AB is a diameter, the angle in the semicircle is \(90^\circ\). So, \(\angle ACB = 90^\circ\).
Step 4: In triangle \(ABC\): \(\angle BAC = 30^\circ\), \(\angle ACB = 90^\circ\). Therefore, \(\angle ABC = 60^\circ\).
Step 5: From Step 2, we know \(\angle ACD = \angle ABC = 60^\circ\).
Step 6: Look at triangle \(BCD\): It has angles \(\angle BCD = 60^\circ\) and \(\angle CBA = 60^\circ\). Therefore, triangle \(BCD\) is isosceles with sides \(BC = BD\).
Final Result: Hence, we have proved that \(BC = BD\).