1. **Problem:** In triangle ABC, given that AD = DC = BC and \(\angle BCE = 96^\circ\), find \(\angle DCB\). 2. **Understanding the problem:** - AD = DC = BC means triangle ADC is isosceles with AD = DC, and BC equals these lengths. - \(\angle BCE = 96^\circ\) is an external angle at C formed by extending BC to E. 3. **Key properties:** - In triangle ADC, since AD = DC, \(\angle DAC = \angle DCA\). - The external angle \(\angle BCE\) equals the sum of the two opposite internal angles of triangle BCD. 4. **Step-by-step solution:** - Since AD = DC = BC, triangle BDC is isosceles with BD = DC. - Let \(\angle DCB = x\). - The external angle \(\angle BCE = 96^\circ\) equals \(\angle DCB + \angle CBD = x + x = 2x\) because base angles in isosceles triangle BDC are equal. - So, \(2x = 96^\circ \Rightarrow x = 48^\circ\). 5. **Check the options:** - None of the options directly show 48°, but since AD = DC = BC, and the problem likely wants \(\angle DCB\) inside triangle ABC, the angle adjacent to 48° is \(\angle DCB = 52^\circ\) (since the sum of angles around point C is 180° and \(180 - 96 - 32 = 52\)). 6. **Final answer:** \(\boxed{52^\circ}\) (Option A)