5.如图，$AD// BC$，$∠ PAB$的平分线与$∠ CBA$的平分线相交于点$E$，连接$CE$并延长交$AP$于点$D$.求证:$AD+BC=AB$.

5. 证明:如答图,在 AB 上截取 AF=AD,连接 EF.
$\because AE$平分$∠ PAB,\therefore ∠ DAE=∠ FAE.$
在$△ DAE$和$△ FAE$中,$\begin{cases} AD=AF,\\ ∠ DAE=∠ FAE,\\ AE=AE, \end{cases}$
$\therefore △ DAE≌△ FAE(\mathrm{SAS}),\therefore ∠ ADE=∠ AFE.$
$\because AD// BC,\therefore ∠ ADE+∠ C=180°.$
$\because ∠ AFE+∠ EFB=180°,\therefore ∠ EFB=∠ C.$
$\because BE$平分$∠ ABC,\therefore ∠ EBF=∠ EBC.$
在$△ BEF$和$△ BEC$中,$\begin{cases} ∠ EFB=∠ C,\\ ∠ EBF=∠ EBC,\\ BE=BE, \end{cases}$
$\therefore △ BEF≌△ BEC(\mathrm{AAS}),\therefore BC=BF.$
$\because AD=AF,\therefore AD+BC=AF+BF=AB.$

6.(玄武区月考)如图,在五边形ABCDE中,AB=AE,BC+DE=CD,∠ABC+∠AED=180°.
求证:DA平分∠CDE.

6. 证明:如答图,连接 AC,延长 DE 到点 F,使 EF=BC,连接 AF.
$\because BC+DE=CD,EF+DE=DF,\therefore CD=FD.$
$\because ∠ ABC+∠ AED=180°,∠ AEF+∠ AED=180°,$
$\therefore ∠ ABC=∠ AEF.$
在$△ ABC$和$△ AEF$中,$\begin{cases} AB=AE,\\ ∠ ABC=∠ AEF,\\ BC=EF, \end{cases}$
$\therefore △ ABC≌△ AEF(\mathrm{SAS}),\therefore AC=AF.$
在$△ ACD$和$△ AFD$中,$\begin{cases} AC=AF\\ CD=FD,\\ AD=AD \end{cases}$
$\therefore △ ACD≌△ AFD(\mathrm{SSS}),\therefore ∠ ADC=∠ ADF,$
即$DA$平分$∠ CDE.$

7.(建湖县月考)如图,已知BD为∠ABC的平分线,DE⊥BC于点E,且AD=CD.
(1)求证:∠BAD+∠BCD=180°;
(2)试探究线段AB,BC,BE之间的数量关系,并说明理由.

7. (1)证明:如答图,过点 D 作$DF⊥ BA$于点 F.
$\because BD$为$∠ ABC$的平分线,$\therefore ∠ EBD=∠ FBD.$
$\because DE⊥ BC,DF⊥ AB,\therefore ∠ BED=∠ BFD=90°.$
在$△ BDE$和$△ BDF$中,$\begin{cases} ∠ EBD=∠ FBD,\\ ∠ BED=∠ BFD,\\ BD=BD, \end{cases}$
$\therefore △ BDE≌△ BDF(\mathrm{AAS}),\therefore DE=DF.$
在$\mathrm{Rt}△ ADF$和$\mathrm{Rt}△ CDE$中,$\begin{cases} AD=CD,\\ DF=DE, \end{cases}$
$\therefore \mathrm{Rt}△ ADF≌\mathrm{Rt}△ CDE(\mathrm{HL}),$
$\therefore ∠ DAF=∠ DCE,AF=CE.$
$\because ∠ DAF+∠ BAD=180°,\therefore ∠ BAD+∠ BCD=180°.$
(2)解:$AB+BC=2BE$.理由如下:
由(1)可知$△ BDF≌△ BDE$,则$BF=BE$,
$\therefore AB+BC=BF-AF+BE+EC=2BE.$

8.如图,在△ABC中,AD平分∠BAC,且BD=CD.求证:∠B=∠C.

8. 证明:如答图,过点 D 作$DE⊥ AB,DF⊥ AC$,则$∠ DEA=∠ DFA=90°$.
$\because AD$平分$∠ BAC,\therefore ∠ BAD=∠ CAD.$
又$\because ∠ DEA=∠ DFA,AD=AD,$
$\therefore △ ADE≌△ ADF(\mathrm{AAS}),\therefore DE=DF.$
又$\because BD=CD,\therefore \mathrm{Rt}△ DBE≌\mathrm{Rt}△ DCF(\mathrm{HL}),$
$\therefore ∠ B=∠ C.$