6. 如图21-29，E，F是正方形ABCD的对角线BD上的两点，且BE=DF.
(1) 求证：$△ ABE ≌ △ CDF$；
(2) 若$AB=3\sqrt{2}$，$BE=2$，求四边形AECF的面积.

图21-29

6. (1)$\because$ 四边形$ABCD$为正方形,$\therefore CD=AB,∠ ABE=∠ CDF=45°$. 又$\because BE=DF,\therefore △ ABE≌ △ CDF(\mathrm{SAS})$.
(2)如图,连接$AC$,交$BD$于点$O. \because$ 四边形$ABCD$是正方形,$\therefore AC⊥ BD,AO=CO,DO=BO$. 又$\because DF=BE,\therefore OE=OF. \therefore$ 四边形$AECF$是平行四边形. $\because AC⊥ EF,\therefore$ 四边形$AECF$是菱形. $\because$ 在正方形$ABCD$中,$AB=3\sqrt{2},\therefore AC=BD=6. \because BE=DF=2,\therefore EF=2. \therefore$ 四边形$AECF$的面积为$\dfrac{1}{2}AC· EF=\dfrac{1}{2}×6×2=6$.

7. 如图21-30,在$□ ABCD$中,O是CD的中点,连接AO并延长,交BC的延长线于点E.

图21-30
(1)求证:$△ AOD≌ △ EOC$.
(2)连接AC,DE,当$∠B = ∠AEB =$时,四边形ACED是正方形.请说明理由.

7. (1)$\because$ 四边形$ABCD$是平行四边形, $\therefore AD// BC. \therefore ∠ ADC = ∠ OCE, ∠ DAO = ∠ OEC$. 又$\because OC=OD,\therefore △ AOD≌ △ EOC$.
(2)$45°$
理由如下:$\because △ AOD≌ △ EOC,\therefore OA=OE$. 又$\because OC=OD,\therefore$ 四边形$ACED$是平行四边形. $\because ∠ B=∠ AEB=45°,\therefore AB=AE,∠ BAE=90°$. $\because$ 四边形$ABCD$是平行四边形,$\therefore AB// CD,AB=CD. \therefore ∠ COE=∠ BAE=90°. \therefore □ ACED$是菱形. $\because AB=AE,AB=CD,\therefore AE=CD. \therefore$ 菱形$ACED$是正方形. $\therefore$ 四边形$ACED$是正方形.