《2025年千里马单元测试卷八年级数学下册人教版》

6. 如图,四边形ABDE和四边形ACFG都是正方形,CE与BG交于点M,点M在△ABC的外部.
(1)求证:BG = CE;
(2)求证:CE⊥BG;
(3)求∠AME的度数.

答案：

(1)证明:在正方形ABDE和正方形ACFG中,AB = AE,AG = AC,∠BAE = ∠CAG = 90°,
∴∠BAE + ∠EAG = ∠CAG + ∠EAG,
即∠BAG = ∠CAE.在△ABG和△AEC中,
$\begin{cases}AB = AE \\\angle BAG=\angle EAC \\AG = AC\end{cases}$
∴△ABG≌△AEC(SAS),
∴BG = CE.
(2)证明:
∵△ABG≌△AEC,
∴∠ACE = ∠AGB,
∴∠MCF + ∠MGF = ∠ACF + ∠AGF = 90° + 90° = 180°,
∴∠CMG = 360° - (∠MCF + ∠MGF + ∠F)=360° - (180° + 90°)=90°,
∴BG⊥CE.
(3)解:如答图,过点A作AP⊥BG,AQ⊥CE,垂足分别为P,Q.

∵△ABG≌△AEC,
∴∠PBA = ∠QEA,AB = AE.
又
∵∠BPA = ∠EQA = 90°,
∴△PBA≌△QEA,
∴AP = AQ,
∴MA平分∠BMC,由
(2)可知∠BMC = 90°,
∴∠AMC = $\frac{1}{2}$∠BMC = 45°,
∴∠AME = 135°.