6. 新考法分类讨论法 如图，已知正方形ABCD的边长为2，对角线AC，BD相交于点O，将△OBC绕点B逆时针旋转得到△O'BC'，当O'，C'，D三点共线时，O'A的长为。

6. $\sqrt{3}-1$或$\sqrt{3}+1$ 【点拨】$\because △ O'C'B$，$△ ABD$是等腰直角三角形，$\therefore O'B:BC'=AB:BD=1:\sqrt{2}$。①当$O'C'$在$BD$的上方时，$\because ∠ ABO'+∠ ABC'=∠ DBC'+∠ ABC'=45°$，$\therefore ∠ ABO' = ∠ DBC'$。$\therefore △ ABO' ∽ △ DBC'$。$\therefore AO':DC'=AB:BD$。$\because BC'=BC=2$，$\therefore O'C'=BO'=\frac{\sqrt{2}}{2}BC'=\sqrt{2}$，$BD=\sqrt{2}BC=2\sqrt{2}$。$\because DO'^2=BD^2-BO'^2$，$\therefore DO'^2=(2\sqrt{2})^2-(\sqrt{2})^2$。$\therefore DO'=\sqrt{6}$。$\therefore DC'=\sqrt{6}-\sqrt{2}$。$\therefore AO':(\sqrt{6}-\sqrt{2})=1:\sqrt{2}$。$\therefore AO'=\sqrt{3}-1$；②当$O'C'$在$BD$的下方时，如图。同理可证$△ ABO' ∽ △ DBC'$，$\therefore AO':DC'=AB:BD$。$\because BC'=BC=2$，$\therefore O'C'=BO'=\frac{\sqrt{2}}{2}BC'=\sqrt{2}$，$BD=\sqrt{2}BC=2\sqrt{2}$。$\because DO'^2=BD^2-BO'^2$，$\therefore DO'^2=(2\sqrt{2})^2-(\sqrt{2})^2$。$\therefore DO'=\sqrt{6}$。$\therefore DC'=\sqrt{6}+\sqrt{2}$。$\therefore AO':(\sqrt{6}+\sqrt{2})=1:\sqrt{2}$。$\therefore AO'=\sqrt{3}+1$。综上所述，$AO'$的长为$\sqrt{3}-1$或$\sqrt{3}+1$。

7. 如图,在等腰直角三角形$ABC$中,$AC=BC$,过点$C$作射线$CP // AB$,$D$为射线$CP$上一点,$E$在边$BC$上(不与点$B$,$C$重合),且$∠ DAE=45°$,连结$DE$交$AC$于点$O$.
(1)求证:$△ ADE ∽ △ ACB$;
(2)如果$△ COD$与$△ BEA$相似,求$CE:BE$的值.

7. (1)【证明】由题意知$∠ CAB=∠ B=45°$。$\because ∠ DAE=45°$，$\therefore ∠ CAD=∠ BAE$。$\because CP// AB$，$\therefore ∠ ACD=∠ CAB=45°$。$\therefore ∠ ACD=∠ B$。$\therefore △ ACD ∽ △ ABE$。$\therefore \frac{AD}{AE}=\frac{AC}{AB}$，即$\frac{AD}{AC}=\frac{AE}{AB}$。又$\because ∠ DAE=∠ CAB=45°$，$\therefore △ ADE ∽ △ ACB$。
(2)【解】在$△ COD$与$△ BEA$中，$∠ DCO=∠ B=45°$。又$\because ∠ DOC$与$∠ AEB$均为钝角，$\therefore$ 如果$△ COD$与$△ BEA$相似，只能是$△ COD ∽ △ BEA$。$\therefore ∠ CDO=∠ BAE$。由(1)易知$∠ AED=45°$，$\therefore ∠ AEC=∠ AED+∠ DEC=45°+∠ DEC$。又$\because ∠ AEC=∠ B+∠ BAE=45°+∠ BAE$，$\therefore ∠ DEC=∠ BAE$。$\therefore ∠ CDO=∠ BAE=∠ DEC$。$\therefore CE=CD$。由(1)知$△ ACD ∽ △ ABE$，$\therefore \frac{DC}{BE}=\frac{AC}{AB}=\frac{AC}{\sqrt{2}AC}=\frac{\sqrt{2}}{2}$。$\therefore \frac{CE}{BE}=\frac{\sqrt{2}}{2}$，即$CE:BE=\sqrt{2}:2$。