1. 如图,P为$△ABC$内一点,$AB= BC,∠ABC= 90^{\circ },PA= 2,PB= 4,PC= 6.$
(1)将$△ABP$绕点B顺时针旋转$90^{\circ }$,画出旋转后的三角形;
(2)求$∠APB$的度数.

解: (1) 如图,$\triangle CBP'$即为所求;

(2) 连接$PP'$.$\because \triangle PAB\cong \triangle P'CB$,$∠PBP'=90^{\circ }$,$\therefore BP=BP'$,$∠APB=∠CP'B$,$CP'=AP=2$,$\therefore PP'=\sqrt {2}PB=4\sqrt {2}$,$∠BP'P=45^{\circ }$.$\because P'P^{2}+P'C^{2}=32+4=36=PC^{2}$,$\therefore ∠CP'P=90^{\circ }$,$\therefore ∠CP'B=∠BP'P+∠CP'P=45^{\circ }+90^{\circ }=135^{\circ }$,$\therefore ∠APB=135^{\circ }$.

2. (教材$P_{63}T_{11}$变式)如图,在四边形ABCD中,$AB= AC,∠BAC= 90^{\circ },∠ADC= 45^{\circ }$.若$AD= 4,$$CD= 2$,求BD的长.

解: 将$\triangle ABD$绕点$A$顺时针旋转$90^{\circ }$得到$\triangle ACE$,连接$DE$,$\therefore ∠EAD=90^{\circ }$,$AE=AD$,$\therefore ∠ADE=∠AED=45^{\circ }$.$ED=\sqrt {2}AD=4\sqrt {2}$,$\because ∠ADC=45^{\circ }$,$\therefore ∠EDC=90^{\circ }$,$\therefore CE^{2}=DE^{2}+CD^{2}=(4\sqrt {2})^{2}+2^{2}=36$,$\therefore CE=6$,$\because \triangle ABD\cong \triangle ACE$,$\therefore BD=CE=6$.

3. 如图,D是$△ABC$内一点,$∠BDC= 90^{\circ },BD= CD,AB= 20,AC= 21,AD= \frac {13\sqrt {2}}{2}$,求BC的长.

解: 将$\triangle ADC$绕点$D$顺时针旋转$90^{\circ }$得到$\triangle EDB$,设$BE$交$AC$于点$F$,连接$AE$.$\because \triangle ADC\cong \triangle EDB$,$\therefore BE=AC=21$,$∠DBE=∠DCA$,$\therefore ∠BFC=∠BDC=90^{\circ }$.$\because AD=DE$,$∠ADE=90^{\circ }$,$\therefore AE=\sqrt {2}AD=13$,设$BF=x$,则$EF=21-x$,$\because AB^{2}-BF^{2}=AE^{2}-EF^{2}=AF^{2}$,$\therefore 20^{2}-x^{2}=13^{2}-(21-x)^{2}$,$\therefore x=16$,即$BF=16$,$\therefore AF=\sqrt {AB^{2}-BF^{2}}=\sqrt {20^{2}-16^{2}}=12$,$\therefore CF=9$,$\therefore BC=\sqrt {BF^{2}+CF^{2}}=\sqrt {337}$.