答案：
C 点拨:如答图,连接 AD,DF.
∵$AB = AC$,$\angle BAC = 90^{\circ}$,D 是 BC 边的中点,
∴$DA = DB = DC$,$\angle DAE = \angle C = 45^{\circ}$,$\angle ADC = 90^{\circ}$.
又
∵$AE = CF$,
∴$\triangle AED \cong \triangle CFD(SAS)$.
∴$DE = DF$,$\angle ADE = \angle CDF$.
∴$\angle EDF = \angle EDA + \angle ADF = \angle CDF + \angle ADF = \angle ADC = 90^{\circ}$,$\therefore \triangle DEF$是等腰直角三角形.
∴$EF = \sqrt{2}DE$.故选 C.

答案：
$AB + AD = \sqrt{2}AC$ 点拨:如答图,过点 C 作$CM \perp AB$于点 M,作$CN \perp AD$,交 AD 的延长线于点 N,
则$\angle CMA = \angle CMB = \angle N = \angle BAD = 90^{\circ}$,
∴$\angle MCN = \angle BCD = 90^{\circ}$.
∴$\angle BCM = \angle DCN$.
又
∵$CB = CD$,
∴$\triangle BCM \cong \triangle DCN(AAS)$.
∴$BM = DN$,$CM = CN$,
∴四边形 AMCN 为正方形,
∴$\triangle CAN$是等腰直角三角形.
∴$AM + AN = \sqrt{2}AC$.
∵$AB + AD = AM + BM + AD = AM + AD + DN = AM + AN$,
∴$AB + AD = \sqrt{2}AC$.

答案：
证明:如答图,延长 AB 至点 E,使$BE = AD$,连接 CE.

∵$\angle ABC$与$\angle D$互补,
∴$\angle CBE = 180^{\circ} - \angle ABC = \angle D$.
又
∵$CB = CD$,$BE = DA$,
∴$\triangle ADC \cong \triangle EBC(SAS)$.
∴$CA = CE$,$\angle ACD = \angle BCE$.
∴$\angle ACE = \angle ACB + \angle BCE = \angle ACB + \angle ACD = \angle BCD = 120^{\circ}$.
又
∵$CA = CE$,
∴$\angle CAE = \angle E = 30^{\circ}$.
∴$AE = \sqrt{3}AC$.$\therefore AB + AD = \sqrt{3}AC$.