19. (9分)如图,已知$∠ABC= ∠ADC= 90^{\circ}$,M,N分别是AC,BD的中点,连接MB,MD.
(1)求证:$MN⊥BD$;
(2)若$∠BAD= 45^{\circ}$,判断$△MBD$的形状,并说明理由.

(1)∵∠ABC = ∠ADC = 90°,M,N分别是AC,BD的中点,∴在Rt△ABC中,BM = $\frac{1}{2}$AC,在Rt△ACD中,DM = $\frac{1}{2}$AC,∴BM = DM.又∵N是BD的中点,∴MN⊥BD.
(2)△MBD是等腰直角三角形.理由:∵M是AC的中点,∴AM = $\frac{1}{2}$AC = BM,∴∠BAM = ∠ABM,∴∠BMC = 2∠BAM.同理可得∠DMC = 2∠DAM.又∵∠BAD = 45°,∴∠BMD = ∠BMC + ∠DMC = 2(∠BAM + ∠DAM) = 2∠BAD = 90°.又∵BM = DM,∴△MBD是等腰直角三角形.

20. (8分)观察与类比:
(1)如图①,在$△ABC$中,$∠ACB= 90^{\circ}$,点D在$△ABC$外,连接AD,作$DE⊥AB$于点E,交BC于点F,$AD= AB,AE= AC$,连接AF,求证:$DF= BC+CF$;
(2)如图②,$AB= AD,AC= AE,∠ACB= ∠AED= 90^{\circ}$,延长BC交DE于点F,写出DF,BC,CF之间的数量关系,并证明你的结论.

(1)∵DE⊥AB,∠ACB = 90°,∴∠AED = ∠AEF = ∠ACB = 90°.在Rt△ACF与Rt△AEF中,{AC = AE,AF = AF,∴Rt△ACF≌Rt△AEF(HL),∴CF = EF.在Rt△ADE与Rt△ABC中,{AD = AB,AE = AC,∴Rt△ADE≌Rt△ABC(HL),∴DE = BC.∵DF = DE + EF,∴DF = BC + CF.
(2)BC = CF + DF.证明如下:连接AF,在Rt△ABC与Rt△ADE中,{AB = AD,AC = AE,∴Rt△ABC≌Rt△ADE(HL),∴BC = DE.∵∠ACB = 90°,∴∠ACF = 90° = ∠AED.在Rt△ACF与Rt△AEF中,{AC = AE,AF = AF,∴Rt△ACF≌Rt△AEF(HL),∴CF = EF.∵DE = EF + DF,∴BC = CF + DF.

21. (10分)下面是某数学兴趣小组探究问题的片段,请仔细阅读,并完成任务.题目背景:在$Rt△ABC$中,$AC= BC,∠ACB= 90^{\circ}$,点D在AB上.
(1)作图探讨:在$Rt△ABC$外侧,以BC为边作$△CBE\cong △CAD$.
小明:如图①,分别以点B,C为圆心,以AD,CD长为半径画弧交于点E,连接BE,CE,则$△CBE$即为所求作的三角形.

小军:如图②,分别过点B,C作AB,CD的垂线,两条垂线相交于点E,则$△CBE$即为所求作的三角形.
填空:小明得出$△CBE\cong △CAD$的依据是____,小军得出$△CBE\cong △CAD$的依据是____.(填序号)
①SSS ②SAS ③ASA ④AAS
(2)测量发现:如图③,在(1)中$△CBE\cong △CAD$的条件下,连接AE.兴趣小组用几何画板测量发现$△CAE和△CDB$的面积相等.为了证明这个发现,尝试延长线段AC至F点,使$CF= CA$,连接EF.请你完成证明过程.

(1)①③
(2)由题图③,得CE是△AEF的中线,∴S△ACE = S△EFC.∵∠ACB = 90°,∴∠BCF = 90°.∵△CBE≌△CAD,∴CE = CD,∠ECB = ∠DCA,∴90° - ∠ECB = 90° - ∠DCA,即∠ECF = ∠DCB.又∵AC = BC,AC = CF,∴CF = CB.在△ECF和△DCB中,{CF = CB,∠ECF = ∠DCB,CE = CD,∴△ECF≌△DCB(SAS).∴S△ECF = S△DCB,∴S△CAE = S△CDB.