\(AB\) is a diameter of a circle and \(AC\) is a chord such that \(\angle BAC = 30^\circ\). If the tangent at \(C\) meets \(AB\) produced at \(D\), then \(BC = BD\). State True/False and justify.
True.
Step 1: We are given that \(AB\) is a diameter of the circle.
From the circle theorem, the angle made by a diameter at the circumference is always a right angle (\(90^\circ\)).
Therefore, in triangle \(ABC\), \(\angle ACB = 90^\circ\).
Step 2: It is also given that \(\angle BAC = 30^\circ\).
Since the sum of angles in a triangle is \(180^\circ\):
\(\angle BAC + \angle ABC + \angle ACB = 180^\circ\)
\(30^\circ + \angle ABC + 90^\circ = 180^\circ\)
\(\angle ABC = 60^\circ\).
Step 3: Now look at the tangent at point \(C\). By the Tangent–Chord Theorem (also called Alternate Segment Theorem):
The angle between the tangent at \(C\) and the chord \(BC\) is equal to the angle in the opposite arc, which is \(\angle BAC\).
So, \(\angle BCD = \angle BAC = 30^\circ\).
Step 4: Now consider triangle \(BCD\):
- \(\angle BCD = 30^\circ\) (from Step 3)
- \(\angle CBD = 60^\circ\) (this is the same as \(\angle ABC\) from Step 2, since point D lies on the extension of AB).
Thus in triangle \(BCD\), the two angles are:
\(\angle CBD = 60^\circ\), \(\angle BCD = 30^\circ\).
Step 5: This makes triangle \(BCD\) a right triangle with angles 90°, 60°, and 30°.
In a 30°–60°–90° triangle, the sides opposite 30° and 60° have a fixed ratio. But here, notice something special: the sides opposite equal angles are equal.
In \(\triangle CBD\), angles at \(C\) and \(D\) are equal (each is 30°). So, the sides opposite them are equal.
That means \(BC = BD\).
Final Conclusion: The statement is True. We proved that \(BC = BD\).