答案：
1. 解:如答图,作$AD⊥ BC$于点$D$.设$CD=x$,则$BD=$$14-x$.$\because AB^{2}-BD^{2}=AC^{2}-CD^{2}$,$\therefore 15^{2}-(14-x)^{2}=$$13^{2}-x^{2}$,$\therefore x=5$,即$CD=5$,$\therefore AD=\sqrt{AC^{2}-CD^{2}}=12$,
$\therefore S_{△ ABC}=\frac{1}{2}×14×12=84$.

第1题答图

答案：
2. 解:如答图,过点$O$作$OE⊥ AB$,$OF⊥ AC$,$OG⊥ BC$,$E$,$F$,
$G$为垂足,过点$N$作$ND⊥ AB$于点$D$,连接$OB$,则$OE=$$OF=OG$.设$OE=x$.

第2题答图
$\because BN=MN=5$,$BM=6$,$ND⊥ AB$,$\therefore DB=DM=3$.
$\therefore ND=\sqrt{5^{2}-3^{2}}=4$.
$\because S_{△ BMN}=S_{△ BMO}+S_{△ BNO}$,$\therefore \frac{1}{2}×6· x+\frac{1}{2}×5· x=$$\frac{1}{2}×6×4$.$\therefore x=\frac{24}{11}$.$\therefore OF=\frac{24}{11}$,
即点$O$到$AC$的距离为$\frac{24}{11}$.

答案：
3. (1)证明:如答图①,延长$AB$至点$E$,使$BE=AD$,连接
$CE$.$\because α =β =90°$,$\therefore ∠ BCD+∠ BAD=180°$,
$\therefore ∠ ADC+∠ ABC=180°$,
$\because ∠ EBC+∠ ABC=180°$,$\therefore ∠ ADC=∠ EBC$.
在$△ ADC$和$△ EBC$中,$\begin{cases} AD=EB,\\ ∠ ADC=∠ EBC,\\ CD=CB,\\ \end{cases}$
$\therefore △ ADC≌△ EBC(\mathrm{SAS})$,$\therefore AC=EC$,$∠ ACD=∠ ECB$.
$\because ∠ ACD+∠ ACB=90°$,
$\therefore ∠ ECB+∠ ACB=90°$,即$∠ ACE=90°$,
$\therefore △ ACE$为等腰直角三角形,
$\therefore AE=\sqrt{2}AC$,$\therefore AB+AD=AB+BE=AE=\sqrt{2}AC$.
(2)证明:如答图②,在线段$AB$上截取$BF=AD$,连接
$CF$.设$AB$与$CD$交于点$H$.
$\because ∠ BCD=∠ BAD$,$∠ BHC=∠ DHA$,$\therefore ∠ B=∠ D$.
在$△ CAD$和$△ CFB$中,$\begin{cases} AD=FB,\\ ∠ D=∠ B,\\ CD=CB,\\ \end{cases}$
$\therefore △ CAD≌△ CFB(\mathrm{SAS})$,$\therefore CA=CF$,$∠ ACD=∠ FCB$.
$\because ∠ FCB+∠ DCF=90°$,
$\therefore ∠ ACD+∠ DCF=90°$,即$∠ ACF=90°$,
$\therefore △ ACF$为等腰直角三角形,
$\therefore AF=\sqrt{2}AC$,$\therefore AB-AD=AB-BF=AF=\sqrt{2}AC$.

第3题答图
(3)证明:如答图③,延长$AB$至点$M$,使$BM=AD$,连接
$CM$,过点$C$作$CH⊥ AB$于点$H$.
由题意得$∠ BCD+∠ BAD=180°$,
$\therefore ∠ ADC+∠ ABC=180°$.
$\because ∠ MBC+∠ ABC=180°$,$\therefore ∠ ADC=∠ MBC$.
$\because DC=BC$,$\therefore △ ADC≌△ MBC$,
$\therefore CA=CM$,$∠ ACD=∠ MCB$,$\therefore ∠ ACM=120°$,
$\therefore ∠ CAH=30°$,$\therefore CH=\frac{1}{2}AC$.
$\therefore AH=\sqrt{AC^{2}-CH^{2}}=\frac{\sqrt{3}}{2}AC$,$\therefore AM=2AH=\sqrt{3}AC$,
$\therefore AB+AD=AB+BM=AM=\sqrt{3}AC$.
(4)解:如答图④,在线段$AB$上截取$BN=AD$,连接$CN$.
由(2)(3)可知,$AB-AD=AB-BN=AN=\sqrt{3}AC$.