8. 如图,在长方形ABCD中,$AB=CD=6cm,BC=10cm$,点P从点B出发,以2 cm/s的速度沿BC向点C运动.当$AP=PC$时,点Q从点C出发,沿CD向点D运动,且始终满足$AP=PQ$.设点P的运动时间为t s,当$t=$
2或2.5
时,以P,C,Q为顶点的三角形与$△ABP$全等.答案：2或2.5
9. (14分)(2024·鼓楼区一模)如图,已知两个滑梯BC和EF的倾斜角$∠ABC$和$∠DFE$互为余角(即$∠ABC+∠DFE=90^{\circ }$),且左边滑梯的高度AC与右边滑梯水平方向的长度DF相等,且$AC⊥BF,ED⊥BF$.小明说:“只要量出左侧滑梯水平方向的长度AB就可以知道右侧滑梯的高度DE了.”他的说法正确吗? 请你说明理由.
解：他的说法
$ \because AC \perp BF $，$ ED \perp BF $，$ \therefore \angle BAC = \angle EDF = 90^\circ $，
$ \therefore \angle ABC + \angle BCA = 90^\circ $.
又$ \angle ABC + \angle DFE = 90^\circ $，$ \therefore \angle BCA = \angle DFE $，
在$ \triangle BAC $与$ \triangle EDF $中，$ \left\{ \begin{array} { l } { \angle BAC = \angle EDF, } \\ { AC = DF, } \\ { \angle BCA = \angle EFD, } \end{array} \right. $
$ \therefore \triangle BAC \cong \triangle EDF ( \text { ASA } ) $，$ \therefore AB = DE $.
解：他的说法
正确
. 理由如下：$ \because AC \perp BF $，$ ED \perp BF $，$ \therefore \angle BAC = \angle EDF = 90^\circ $，
$ \therefore \angle ABC + \angle BCA = 90^\circ $.
又$ \angle ABC + \angle DFE = 90^\circ $，$ \therefore \angle BCA = \angle DFE $，
在$ \triangle BAC $与$ \triangle EDF $中，$ \left\{ \begin{array} { l } { \angle BAC = \angle EDF, } \\ { AC = DF, } \\ { \angle BCA = \angle EFD, } \end{array} \right. $
$ \therefore \triangle BAC \cong \triangle EDF ( \text { ASA } ) $，$ \therefore AB = DE $.
答案：解：他的说法正确. 理由如下：
$ \because AC \perp BF $，$ ED \perp BF $，$ \therefore \angle BAC = \angle EDF = 90^\circ $，
$ \therefore \angle ABC + \angle BCA = 90^\circ $.
又$ \angle ABC + \angle DFE = 90^\circ $，$ \therefore \angle BCA = \angle DFE $，
在$ \triangle BAC $与$ \triangle EDF $中，$ \left\{ \begin{array} { l } { \angle BAC = \angle EDF, } \\ { AC = DF, } \\ { \angle BCA = \angle EFD, } \end{array} \right. $
$ \therefore \triangle BAC \cong \triangle EDF ( \text { ASA } ) $，$ \therefore AB = DE $.
$ \because AC \perp BF $，$ ED \perp BF $，$ \therefore \angle BAC = \angle EDF = 90^\circ $，
$ \therefore \angle ABC + \angle BCA = 90^\circ $.
又$ \angle ABC + \angle DFE = 90^\circ $，$ \therefore \angle BCA = \angle DFE $，
在$ \triangle BAC $与$ \triangle EDF $中，$ \left\{ \begin{array} { l } { \angle BAC = \angle EDF, } \\ { AC = DF, } \\ { \angle BCA = \angle EFD, } \end{array} \right. $
$ \therefore \triangle BAC \cong \triangle EDF ( \text { ASA } ) $，$ \therefore AB = DE $.
10. (18分)(2023·兴化月考)如图,$AC⊥BC,DC⊥EC,AC=BC,DC=EC$.AE与BD有怎样的数量关系和位置关系? 试证明你的结论.
解：$ AE $
证明：$ \because AC \perp BC $，$ DC \perp EC $，$ \therefore \angle ACB = \angle DCE = 90^\circ $，
$ \because \angle ACD = \angle ACD $，
$ \therefore \angle DCB = \angle ECA $.
在$ \triangle DCB $和$ \triangle ECA $中，$ \left\{ \begin{array} { l } { BC = AC, } \\ { \angle DCB = \angle ECA, } \\ { CD = CE, } \end{array} \right. $
$ \therefore \triangle DCB \cong \triangle ECA $
$ \therefore \angle A = \angle B $，$ BD = AE $.
设$ BD $与$ AC $交于点$ N $.
$ \because \angle AND = \angle BNC $，$ \angle B + \angle BNC = 90^\circ $，
$ \therefore \angle A + \angle AND = 90^\circ $，
$ \therefore BD \perp AE $.
解：$ AE $
=
$ BD $，$ AE $⊥
$ BD $.证明：$ \because AC \perp BC $，$ DC \perp EC $，$ \therefore \angle ACB = \angle DCE = 90^\circ $，
$ \because \angle ACD = \angle ACD $，
$ \therefore \angle DCB = \angle ECA $.
在$ \triangle DCB $和$ \triangle ECA $中，$ \left\{ \begin{array} { l } { BC = AC, } \\ { \angle DCB = \angle ECA, } \\ { CD = CE, } \end{array} \right. $
$ \therefore \triangle DCB \cong \triangle ECA $
(SAS)
，$ \therefore \angle A = \angle B $，$ BD = AE $.
设$ BD $与$ AC $交于点$ N $.
$ \because \angle AND = \angle BNC $，$ \angle B + \angle BNC = 90^\circ $，
$ \therefore \angle A + \angle AND = 90^\circ $，
$ \therefore BD \perp AE $.
答案：解：$ AE = BD $，$ AE \perp BD $.
证明：$ \because AC \perp BC $，$ DC \perp EC $，$ \therefore \angle ACB = \angle DCE = 90^\circ $，
$ \because \angle ACD = \angle ACD $，
$ \therefore \angle DCB = \angle ECA $.
在$ \triangle DCB $和$ \triangle ECA $中，$ \left\{ \begin{array} { l } { BC = AC, } \\ { \angle DCB = \angle ECA, } \\ { CD = CE, } \end{array} \right. $
$ \therefore \triangle DCB \cong \triangle ECA ( \text { SAS } ) $，
$ \therefore \angle A = \angle B $，$ BD = AE $.
设$ BD $与$ AC $交于点$ N $.
$ \because \angle AND = \angle BNC $，$ \angle B + \angle BNC = 90^\circ $，
$ \therefore \angle A + \angle AND = 90^\circ $，
$ \therefore BD \perp AE $.
证明：$ \because AC \perp BC $，$ DC \perp EC $，$ \therefore \angle ACB = \angle DCE = 90^\circ $，
$ \because \angle ACD = \angle ACD $，
$ \therefore \angle DCB = \angle ECA $.
在$ \triangle DCB $和$ \triangle ECA $中，$ \left\{ \begin{array} { l } { BC = AC, } \\ { \angle DCB = \angle ECA, } \\ { CD = CE, } \end{array} \right. $
$ \therefore \triangle DCB \cong \triangle ECA ( \text { SAS } ) $，
$ \therefore \angle A = \angle B $，$ BD = AE $.
设$ BD $与$ AC $交于点$ N $.
$ \because \angle AND = \angle BNC $，$ \angle B + \angle BNC = 90^\circ $，
$ \therefore \angle A + \angle AND = 90^\circ $，
$ \therefore BD \perp AE $.
11. (20分)如图,四边形ABCD中,$AB=BC=2CD,AB// CD,∠C=90^{\circ }$,E是BC的中点,AE与BD相交于点F,连接DE.
(1)求证:$△ABE\cong △BCD$;
证明：$ \because AB // CD ，$$ \therefore \angle ABE + \angle C = 180^\circ .$
$ \because \angle C = 90^\circ ，$$ \therefore \angle ABE = 90^\circ = \angle C .$
$ \because E $是 BC 的中点，$ \therefore BC = 2BE ，$
$ \because BC = 2CD ，$$ \therefore BE = CD .$
$ \therefore \triangle ABE \cong \triangle BCD ( \text {
(2)判断线段AE与BD的数量关系及位置关系,并说明理由;
解：AE
由(1)得$ \triangle ABE \cong \triangle BCD ，$
$ \therefore AE = BD ，$$ \angle BAE = \angle CBD ，$
$ \because \angle ABF + \angle CBD = 90^\circ ，$
$ \therefore \angle ABF + \angle BAE = 90^\circ ，$
$ \therefore \angle AFB = 90^\circ ，$$ \therefore AE \perp BD .$
(3)若$CD=1$,试求$△AED$的面积.
解：$ \because \triangle ABE \cong \triangle BCD ，$
$ \therefore BE = CD = 1 ，$ AB = BC = 2CD = 2 ，
$ \therefore CE = BC - BE = 1 ，$
$ \therefore \triangle AED $的面积 = 梯形 ABCD 的面积$ - \triangle ABE $的面积$ -\triangle CDE$的面积$=\frac{1}{2}×(1 + 2)×2-\frac{1}{2}×2×1-\frac{1}{2}×1×1=
(1)求证:$△ABE\cong △BCD$;
证明：$ \because AB // CD ，$$ \therefore \angle ABE + \angle C = 180^\circ .$
$ \because \angle C = 90^\circ ，$$ \therefore \angle ABE = 90^\circ = \angle C .$
$ \because E $是 BC 的中点，$ \therefore BC = 2BE ，$
$ \because BC = 2CD ，$$ \therefore BE = CD .$
$ \therefore \triangle ABE \cong \triangle BCD ( \text {
SAS
} ) .$(2)判断线段AE与BD的数量关系及位置关系,并说明理由;
解：AE
=
BD ，AE ⊥
BD . 理由如下：由(1)得$ \triangle ABE \cong \triangle BCD ，$
$ \therefore AE = BD ，$$ \angle BAE = \angle CBD ，$
$ \because \angle ABF + \angle CBD = 90^\circ ，$
$ \therefore \angle ABF + \angle BAE = 90^\circ ，$
$ \therefore \angle AFB = 90^\circ ，$$ \therefore AE \perp BD .$
(3)若$CD=1$,试求$△AED$的面积.
解：$ \because \triangle ABE \cong \triangle BCD ，$
$ \therefore BE = CD = 1 ，$ AB = BC = 2CD = 2 ，
$ \therefore CE = BC - BE = 1 ，$
$ \therefore \triangle AED $的面积 = 梯形 ABCD 的面积$ - \triangle ABE $的面积$ -\triangle CDE$的面积$=\frac{1}{2}×(1 + 2)×2-\frac{1}{2}×2×1-\frac{1}{2}×1×1=
\frac{3}{2}
$答案：(1) 证明：$ \because AB // CD $，$ \therefore \angle ABE + \angle C = 180^\circ $.
$ \because \angle C = 90^\circ $，$ \therefore \angle ABE = 90^\circ = \angle C $.
$ \because E $是$ BC $的中点，$ \therefore BC = 2BE $，
$ \because BC = 2CD $，$ \therefore BE = CD $.
在$ \triangle ABE $和$ \triangle BCD $中，$ \left\{ \begin{array} { l } { BE = CD, } \\ { \angle ABE = \angle C, } \\ { AB = BC, } \end{array} \right. $
$ \therefore \triangle ABE \cong \triangle BCD ( \text { SAS } ) $.
(2) 解：$ AE = BD $，$ AE \perp BD $. 理由如下：
由(1)得$ \triangle ABE \cong \triangle BCD $，
$ \therefore AE = BD $，$ \angle BAE = \angle CBD $，
$ \because \angle ABF + \angle CBD = 90^\circ $，
$ \therefore \angle ABF + \angle BAE = 90^\circ $，
$ \therefore \angle AFB = 90^\circ $，$ \therefore AE \perp BD $.
(3) 解：$ \because \triangle ABE \cong \triangle BCD $，
$ \therefore BE = CD = 1 $，$ AB = BC = 2CD = 2 $，
$ \therefore CE = BC - BE = 1 $，
$ \therefore \triangle AED $的面积$ = $梯形$ ABCD $的面积$ - \triangle ABE $的面积$ -
$\triangle CDE$的面积$=\frac{1}{2}×(1 + 2)×2-\frac{1}{2}×2×1-\frac{1}{2}×1×1=\frac 32$
$ \because \angle C = 90^\circ $，$ \therefore \angle ABE = 90^\circ = \angle C $.
$ \because E $是$ BC $的中点，$ \therefore BC = 2BE $，
$ \because BC = 2CD $，$ \therefore BE = CD $.
在$ \triangle ABE $和$ \triangle BCD $中，$ \left\{ \begin{array} { l } { BE = CD, } \\ { \angle ABE = \angle C, } \\ { AB = BC, } \end{array} \right. $
$ \therefore \triangle ABE \cong \triangle BCD ( \text { SAS } ) $.
(2) 解：$ AE = BD $，$ AE \perp BD $. 理由如下：
由(1)得$ \triangle ABE \cong \triangle BCD $，
$ \therefore AE = BD $，$ \angle BAE = \angle CBD $，
$ \because \angle ABF + \angle CBD = 90^\circ $，
$ \therefore \angle ABF + \angle BAE = 90^\circ $，
$ \therefore \angle AFB = 90^\circ $，$ \therefore AE \perp BD $.
(3) 解：$ \because \triangle ABE \cong \triangle BCD $，
$ \therefore BE = CD = 1 $，$ AB = BC = 2CD = 2 $，
$ \therefore CE = BC - BE = 1 $，
$ \therefore \triangle AED $的面积$ = $梯形$ ABCD $的面积$ - \triangle ABE $的面积$ -
$\triangle CDE$的面积$=\frac{1}{2}×(1 + 2)×2-\frac{1}{2}×2×1-\frac{1}{2}×1×1=\frac 32$