4. 如图，Rt△ACB中，∠ACB=90°，△ABC的角平分线AD，BE相交于点P，过点P作PF⊥AD交BC的延长线于点F，交AC于点H.求证：
(1) $△ ABP ≌ △ FBP$；
(2) $AH+BD=AB$.

4. (1)
∵ $∠ACB=90°, \therefore ∠CAB+∠CBA=90°.$又
∵ AD,BE 分别平分 $∠BAC,∠ABC, \therefore ∠BAD+∠ABE=\frac{1}{2}(∠CAB+∠CBA)=45°, \therefore ∠APB=135°, \therefore ∠BPD=45°.$ 又
∵ $PF ⊥ AD, \therefore ∠FPB=90°+45°=135°, \therefore ∠APB = ∠FPB.$ 在 $△ ABP$ 和 $△ FBP$ 中, $\begin{cases}∠ABP=∠FBP,\\BP=BP,\\∠APB=∠FPB,\end{cases}$ $\therefore △ ABP ≌ △ FBP(\mathrm{ASA}).$
(2)
∵ $△ ABP ≌ △ FBP, \therefore PA = PF, ∠BAP = ∠F.$
∵ $∠BAP = ∠CAD, \therefore ∠F = ∠CAD.$ 在 $△ APH$ 和 $△ FPD$ 中, $\begin{cases}∠APH=∠FPD,\\PA=PF,\\∠PAH=∠PFD,\end{cases}$ $\therefore △ APH ≌ △ FPD(\mathrm{ASA}), \therefore AH=FD.$ 又
∵ $AB=FB, \therefore AB=FD+BD=AH+BD.$

5. 如图，直角$△ ABC$中，$AC=BC$，$AD$平分$∠ BAC$交$BC$于点$D$，$CE ⊥ AD$交$AD$于点$F$，交$AB$于点$E$.求证：$AD=2DF+CE$.

5. 如图,在 AF 上截取 $FG = DF$,连结 CG,则 $DG=2DF.$
∵ $∠ACB=90°, \therefore ∠DCF+∠ACF=90°.$又
∵ $CF ⊥ AD, \therefore ∠ACF+∠CAF=90°, \therefore ∠DCF = ∠CAF.$
∵ AD 平分 $∠CAE, \therefore ∠CAF = ∠EAF.$
∵ $DF=FG, CF ⊥ DG, \therefore CD = CG, \therefore ∠CDG = ∠CGD.$
∵ $∠DGC = ∠GAC+∠ACG, ∠ADC = ∠B+∠BAD, \therefore ∠B = ∠ACG.$又
∵ $AC=BC, \therefore △ ACG ≌ △ CBE(\mathrm{ASA}), \therefore AG=CE, \therefore AD=AG+DG=CE+2DF.$

6. 如图，$∠ BAD=∠ CAE=90°,AB=AD,AE=AC,AF⊥ CB$,垂足为$F$.
(1)求证:$△ BAC≌△ DAE$;
(2)求证:$CD=2BF+DE$.

6. (1)
∵ $∠BAD=∠CAE=90°, \therefore ∠BAC+∠CAD=90°, ∠CAD+∠DAE = 90°, \therefore ∠BAC = ∠DAE.$ 在 $△ BAC$ 和 $△ DAE$ 中, $\begin{cases}AB=AD,\\∠BAC=∠DAE,\\AC=AE,\end{cases}$ $\therefore △ BAC ≌ △ DAE(\mathrm{SAS}).$
(2)如图,延长 BF 到点 G,使得 $FG = FB.$
∵ $AF ⊥ BG, \therefore ∠AFG = ∠AFB = 90°.$在 $△ AFB$ 和 $△ AFG$ 中, $\begin{cases}BF=GF,\\∠AFB=∠AFG,\\AF=AF,\end{cases}$ $\therefore △ AFB ≌ △ AFG(\mathrm{SAS}), \therefore AB = AG, ∠ABF = ∠G.$
∵ $△ BAC ≌ △ DAE, \therefore AB = AD, ∠CBA = ∠EDA, ∠BCA = ∠E, CB = ED, \therefore AG=AD, ∠ABF = ∠CDA, \therefore ∠G = ∠CDA, \therefore ∠GCA = ∠E = ∠DCA = 45°.$在 $△ CGA$ 和 $△ CDA$ 中, $\begin{cases}∠GCA=∠DCA,\\∠CGA=∠CDA,\\AG=AD,\end{cases}$ $\therefore △ CGA ≌ △ CDA(\mathrm{AAS}), \therefore CG=CD.$
∵ $CG = CB+BF+FG = CB+2BF = DE+2BF, \therefore CD=2BF+DE.$