(1)①CF与BD位置关系是垂直、数量关系是相等;(1分)
②当点D在BC的延长线上时①的结论仍成立(如图3).
由正方形ADEF得AD=AF,∠DAF=90°,
∵∠BAC=90°,
∴∠DAF=∠BAC,
∴∠DAB=∠FAC,
又AB=AC,
∴△DAB≌△FAC,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/caef76094b36acafa9714f4d7fd98d1000e99cea?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴CF=BD,
∠ACF=∠ABD.
∵∠BAC=90°,AB=AC,
∴∠ABC=45°,∴∠ACF=45°,
∴∠BCF=∠ACB+∠ACF=90°.即CF⊥BD.(3分)
(2)①画出图形(如图4),判断:(1)中的结论不成立.
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/9345d688d43f87944550919fd11b0ef41ad53ad0?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
②画出图形(如图5),判断:(1)中的结论不成立.(4分)
(3)当∠BCA=45°时,CF⊥BD(如图6).
理由是:过点A作AG⊥AC交BC于点G,
∴AC=AG.
∵∠BCA=45°,
∴∠AGD=45°,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/1ad5ad6eddc451da6859d55ab5fd5266d11632d0?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴△GAD≌△CAF
∴∠ACF=∠AGD=45°.
∠BCF=∠ACB+∠ACF=90°
即CF⊥BD.(5分)
(4)当具备∠BCA=45°时,
过点A作AQ⊥BC交CB的延长线于点Q,(如图7),
∵DE与CF交于点P时,此时点D位于线段CQ上,
∵∠BCA=45°,AC=
2,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/38dbb6fd5266d016e7b64799942bd40734fa35d0?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴由勾股定理可求得AQ=CQ=2.
设CD=x,∴DQ=2-x,
∵∠ADB+∠ADE+∠PDC=180°
且∠ADE=90°,
∴∠ADQ+∠PDC=90°,
又∵在直角△PCD中,∠PDC+∠DPC=90°
∴∠ADQ=∠DPC,
∵∠AQD=∠DCP=90°
∴△AQD∽△DCP,
∴
=,∴
=.
∴CP=-
x
2+x=-
(x-1)
2+
.(7分)
∵0<x≤
,
∴当x=1时,CP有最大值
.(8分)