Answer: A
In triangle \(\triangle\)ABC: \(\angle\)ABC + \(\angle\)ACB + \(\angle\)BAC = 180\(^\circ\) Or, \(2\angle\)ACB + \(\angle\)BAC = 180\(^\circ\) [because AB = AC] Or, \(2\angle\)ACB + \(\angle\)ACD + \(\angle\)ADC = 180\(^\circ\) [since \(\angle\)BAC is the exterior angle of triangle ACD] Or, \(2\angle\)ACB + \(2\angle\)ACD = 180\(^\circ\) [because AC = AD] Or, \(2(\angle\)ACB + \(\angle\)ACD) = 180\(^\circ\) Or, \(2\angle\)BCD = 180\(^\circ\) Therefore, \(\angle\)BCD = 90\(^\circ = \frac{\pi}{2}\) radians
In triangle \(\triangle\)ABC: \(\angle\)ABC + \(\angle\)ACB + \(\angle\)BAC = 180\(^\circ\) Or, \(2\angle\)ACB + \(\angle\)BAC = 180\(^\circ\) [because AB = AC] Or, \(2\angle\)ACB + \(\angle\)ACD + \(\angle\)ADC = 180\(^\circ\) [since \(\angle\)BAC is the exterior angle of triangle ACD] Or, \(2\angle\)ACB + \(2\angle\)ACD = 180\(^\circ\) [because AC = AD] Or, \(2(\angle\)ACB + \(\angle\)ACD) = 180\(^\circ\) Or, \(2\angle\)BCD = 180\(^\circ\) Therefore, \(\angle\)BCD = 90\(^\circ = \frac{\pi}{2}\) radians