Let \(\triangle ABC\) be a triangle with \(AB = AC\) and \(D\) be a point on \(BC\) such that \(\angle BAD = 30^\circ\). If \(E\) is a point on \(AC\) such that \(AD = AE\), then \(\angle CDE\) equals

Let angle B = angle C = x. Angle ADC = x+30 (exterior angle of ABD). Let angle CDE = y; angle AED = angle ADE = x+30-y, and angle AED is exterior to triangle DEC so equals x+y. Hence x+30-y=x+y, y=15 degrees.