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相等的边：$AB=CD$，$AC=CA$，$BC=DA$
相等的角：$\angle BAC = \angle DCA$，$\angle B = \angle D$，$\angle ACB = \angle CAD$
解：$\because \triangle ABC \cong \triangle A'B'C'$，$\therefore \angle B = \angle B' = 51°$，$\angle C = \angle C'$．$\therefore \angle C = 180° - 56° - 51° = 73°$．
解：$\because \triangle ABC \cong \triangle DCB$，$\therefore \angle DBC = \angle ACB = 40°$．$\therefore \angle BEC = 180° - 40° - 40° = 100°$．


相等的边：$AB=CD$，$AC=CA$，$BC=DA$
相等的角：$\angle BAC = \angle DCA$，$\angle B = \angle D$，$\angle ACB = \angle CAD$
解：$\because \triangle ABC \cong \triangle A'B'C'$，$\therefore \angle B = \angle B' = 51°$，$\angle C = \angle C'$．$\therefore \angle C = 180° - 56° - 51° = 73°$．
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