如图所示$triangleABC$中,$angleC=90^{circ},angleA,angleB$的平分线交于$D$点,$DEperpBC$于点$E,DFperpAC$于点F.(1)求证:四边形$CEDF$为正方形;(2)若$AC=12,BC=16$,求$CE$的长.<ImageHere>
(1)直接利用矩形的判定方法以及角平分线的性质得出四边形$CEDF$为正方形;(2)利用三角形面积求法得出$CE$的长.【解析】（1）证明：过点$D$作$DNperpAB$于点$N$,<ImageHere>$becauseangleC=90^{circ},DEperpBC$于点$E,DFperpAC$于点$F$,$therefore$四边形FCED是矩形,又$becauseangleA,angleB$的平分线交于$D$点,$thereforeDF=DE=DN$,$therefore$矩形FCED是正方形;(2)解:$becauseAC=12,BC=16,angleC=90^{circ}$,$thereforeAB=sqrt{AC^{2}+BC^{2}}=sqrt{12^{2}+16^{2}}=20$,$because$四边形$CEDF$为正方形,$thereforeDF=DE=DN$,$thereforeDFtimesAC+DEtimesBC+DNtimesAB=ACtimesBC$,则$EC(AC+BC+AB)=ACtimesBC$,故${EC}=frac{12times16}{12+16+20}=4$.【点睛】此题主要考查了正方形的判定以及三角形面积求法和角平分线的性质等知识,得出$DF=$$DE$是解题关键.