7. (2024·常州一模)如图,在正方形网格中,若$\triangle MNP\cong \triangle MEQ$,则点$Q$可能是$A$,$B$,$C$,$D$四个点中的点______

D
.
答案:D
8. (2024春·工业园区期末)如图,已知$\triangle ABC\cong \triangle DEF$,$\angle C=\angle DFE=90^{\circ}$,先将两个三角形重叠在一起,再将$\triangle DEF$沿$CA$方向平移$2cm$,$AB$,$EF$相交于点$G$.若$BC=8cm$,$GE=3cm$,则阴影部分的面积为______

13
$cm^{2}$.
答案:13
9. 如图,$A$,$D$,$E$三点在同一条直线上,且$\triangle BAD\cong \triangle ACE$.
(1)求证:$BD=DE+CE$;
(2)当$\angle BAC$满足什么条件时,$BD// CE$?

(1)证明:∵△BAD≌△ACE,∴BD = AE,AD = CE,∴BD = AE = AD + DE = CE + DE,
∴BD = DE + CE.
(2)解:∠BAC =
∵△BAD≌△ACE,
∴∠CAE = ∠ABD,∠ADB = ∠AEC.
∵∠BAC = 90°,∴∠BAE + ∠CAE = 90°,
∴∠ABD + ∠BAD = 90°,∴∠ADB = 90°,∴∠BDE = 90°.又∠AEC = ∠ADB = 90°,
∴∠BDE = ∠AEC,∴BD//CE.
(1)求证:$BD=DE+CE$;
(2)当$\angle BAC$满足什么条件时,$BD// CE$?

(1)证明:∵△BAD≌△ACE,∴BD = AE,AD = CE,∴BD = AE = AD + DE = CE + DE,
∴BD = DE + CE.
(2)解:∠BAC =
90°
时,BD//CE;理由:∵△BAD≌△ACE,
∴∠CAE = ∠ABD,∠ADB = ∠AEC.
∵∠BAC = 90°,∴∠BAE + ∠CAE = 90°,
∴∠ABD + ∠BAD = 90°,∴∠ADB = 90°,∴∠BDE = 90°.又∠AEC = ∠ADB = 90°,
∴∠BDE = ∠AEC,∴BD//CE.
答案:(1)证明:∵△BAD≌△ACE,∴BD = AE,AD = CE,∴BD = AE = AD + DE = CE + DE,
∴BD = DE + CE.
(2)解:∠BAC = 90°时,BD//CE;理由:
∵△BAD≌△ACE,
∴∠CAE = ∠ABD,∠ADB = ∠AEC.
∵∠BAC = 90°,∴∠BAE + ∠CAE = 90°,
∴∠ABD + ∠BAD = 90°,∴∠ADB = 90°,∴∠BDE = 90°.又∠AEC = ∠ADB = 90°,
∴∠BDE = ∠AEC,∴BD//CE.
∴BD = DE + CE.
(2)解:∠BAC = 90°时,BD//CE;理由:
∵△BAD≌△ACE,
∴∠CAE = ∠ABD,∠ADB = ∠AEC.
∵∠BAC = 90°,∴∠BAE + ∠CAE = 90°,
∴∠ABD + ∠BAD = 90°,∴∠ADB = 90°,∴∠BDE = 90°.又∠AEC = ∠ADB = 90°,
∴∠BDE = ∠AEC,∴BD//CE.
10. 如图,点$A$,$B$,$C$在同一条直线上,点$E$在$BD$上,且$\triangle ABD\cong \triangle EBC$,$AB=2cm$,$BC=3cm$.
(1)求$DE$的长;
(2)判断$AC$与$BD$的位置关系,并说明理由;
(3)判断$AD$与$CE$的位置关系,并说明理由.

(1)求$DE$的长;
(2)判断$AC$与$BD$的位置关系,并说明理由;
(3)判断$AD$与$CE$的位置关系,并说明理由.

答案:
解:(1)∵△ABD≌△EBC,∴BD = BC = 3cm,BE = AB = 2cm,∴DE = BD - BE = 1cm
(2)DB与AC垂直.
理由:∵△ABD≌△EBC,∴∠ABD = ∠EBC,
又点A,B,C在同一条直线上,
∴∠EBC = 90°,
∴DB与AC垂直.
(3)AD与CE垂直.
理由:如答图,延长CE交AD于点F.
∵△ABD≌△EBC,
∴∠D = ∠C;
∵在△ABD中,∠A + ∠D = 90°,
∴∠A + ∠C = 90°,∴∠AFC = 90°,即CE⊥AD.

解:(1)∵△ABD≌△EBC,∴BD = BC = 3cm,BE = AB = 2cm,∴DE = BD - BE = 1cm
(2)DB与AC垂直.
理由:∵△ABD≌△EBC,∴∠ABD = ∠EBC,
又点A,B,C在同一条直线上,
∴∠EBC = 90°,
∴DB与AC垂直.
(3)AD与CE垂直.
理由:如答图,延长CE交AD于点F.
∵△ABD≌△EBC,
∴∠D = ∠C;
∵在△ABD中,∠A + ∠D = 90°,
∴∠A + ∠C = 90°,∴∠AFC = 90°,即CE⊥AD.
