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对应边：$MQ$与$PN$，$MO$与$PO$，$OQ$与$ON$
对应角：$\angle M$与$\angle P$，$\angle Q$与$\angle N$，$\angle MOQ$与$\angle PON$
解：$\because \triangle ABC \cong \triangle DBC$，$\therefore \angle A = \angle D = 45°$，$\angle ACB = \angle DCB$．又$\because \angle ACB + \angle DCB = \angle ACD = 76°$，$\therefore \angle DCB = \angle ACB = 38°$．$\therefore \angle DBC = 180° - \angle DCB - \angle D = 180° - 38° - 45° = 97°$．
（1）相等．$\because \triangle ABE \cong \triangle ACF$，$\therefore AB = AC$，$AE = AF$，$\therefore AB - AF = AC - AE$，即$BF = CE$．
（2）相等．$\because \triangle ABE \cong \triangle ACF$．$\therefore \angle AEB = \angle AFC$．$\therefore \angle BFO = \angle CEO$．


对应边：$MQ$与$PN$，$MO$与$PO$，$OQ$与$ON$
对应角：$\angle M$与$\angle P$，$\angle Q$与$\angle N$，$\angle MOQ$与$\angle PON$
解：$\because \triangle ABC \cong \triangle DBC$，$\therefore \angle A = \angle D = 45°$，$\angle ACB = \angle DCB$．又$\because \angle ACB + \angle DCB = \angle ACD = 76°$，$\therefore \angle DCB = \angle ACB = 38°$．$\therefore \angle DBC = 180° - \angle DCB - \angle D = 180° - 38° - 45° = 97°$．
（1）相等．$\because \triangle ABE \cong \triangle ACF$，$\therefore AB = AC$，$AE = AF$，$\therefore AB - AF = AC - AE$，即$BF = CE$．
（2）相等．$\because \triangle ABE \cong \triangle ACF$．$\therefore \angle AEB = \angle AFC$．$\therefore \angle BFO = \angle CEO$．
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