《2025年千里马单元测试卷八年级数学下册人教版》

7. 如图,正方形ABCD中,∠EAF的两边分别与边BC,CD交于点E,F,AE,AF分别交BD于点G,H,且∠EAF = 45°.
(1)当∠AEB = 55°时,求∠DAH的度数;
(2)设∠AEB = α,则∠AFD = ______(用含α的代数式表示);
(3)求证:∠AEB = ∠AEF.

答案：

(1)解:
∵四边形ABCD为正方形,
∴∠DAB = ∠ABC = ∠C = ∠ADC = 90°.
当∠AEB = 55°时,
∠EAB = 90° - ∠AEB = 90° - 55° = 35°,
∴∠DAH = 90° - ∠EAF - ∠EAB = 90° - 45° - 35° = 10°.
(2)解:135° - α [解析]由四边形ABCD为正方形可知∠ABE = ∠ADF = ∠BAD = 90°.
∵∠AEB = α,
∴∠EAB = 90° - α,
∴∠DAF = ∠BAD - ∠EAB - ∠EAF = 90° - (90° - α) - 45° = α - 45°,
∴∠AFD = 90° - ∠DAF = 90° - (α - 45°)=135° - α.
(3)证明:将△ADF绕点A顺时针旋转90°得到△ABI,如答图所示,可得E,B,I三点共线,由旋转可知∠DAF = ∠BAI,AF = AI.
∵∠DAF + ∠EAB = 90° - ∠EAF = 45°,
∴∠BAI + ∠EAB = 45° = ∠IAE.在△EAF和△EAI中,
$\begin{cases}AF = AI \\\angle EAF=\angle EAI \\AE = AE\end{cases}$
∴△EAF≌△EAI(SAS),
∴∠AEF = ∠AEB.