1.(2023·苏州期中)如图,在正方形ABCD中,E为AD上一点,连接BE,BE交对角线于点F,连接DF,若∠ABE = 35°,则∠CFD的度数为( )

A. 80°
B. 70°
C. 75°
D. 45°
A. 80°
B. 70°
C. 75°
D. 45°
答案
A
2.(2023·建邺区期末)如图,在平面直角坐标系中,正方形ABCD的边长为2,∠DAO = 60°,则点C的坐标为________.

答案
$(\sqrt{3},1 + \sqrt{3})$
3. 如图,E是正方形ABCD的边BC延长线上的一点,且CE = AC.
(1)求∠E的度数;
(2)若AB = 1,求△ACE的面积.

(1)求∠E的度数;
(2)若AB = 1,求△ACE的面积.
答案
解:(1) $\because$ 四边形 $ABCD$ 为正方形,
$\therefore \angle BCA = \angle ACD = \frac{1}{2}\angle BCD = 45^{\circ}$,
$\therefore \angle ACE = 180^{\circ} - \angle BCA = 180^{\circ} - 45^{\circ} = 135^{\circ}$.
$\because CE = AC$,
$\therefore \angle E = \angle CAE = \frac{1}{2}\times(180^{\circ} - 135^{\circ}) = 22.5^{\circ}$.
(2) $\because$ 四边形 $ABCD$ 为正方形,
$\therefore AB = BC = CD = AD = 1,\angle B = 90^{\circ}$,
$\therefore AC = \sqrt{AB^{2} + BC^{2}} = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$,
$\therefore CE = AC = \sqrt{2}$,
$\therefore S_{\triangle ACE} = \frac{1}{2}CE\cdot AB = \frac{1}{2}\times\sqrt{2}\times1 = \frac{\sqrt{2}}{2}$.
$\therefore \angle BCA = \angle ACD = \frac{1}{2}\angle BCD = 45^{\circ}$,
$\therefore \angle ACE = 180^{\circ} - \angle BCA = 180^{\circ} - 45^{\circ} = 135^{\circ}$.
$\because CE = AC$,
$\therefore \angle E = \angle CAE = \frac{1}{2}\times(180^{\circ} - 135^{\circ}) = 22.5^{\circ}$.
(2) $\because$ 四边形 $ABCD$ 为正方形,
$\therefore AB = BC = CD = AD = 1,\angle B = 90^{\circ}$,
$\therefore AC = \sqrt{AB^{2} + BC^{2}} = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$,
$\therefore CE = AC = \sqrt{2}$,
$\therefore S_{\triangle ACE} = \frac{1}{2}CE\cdot AB = \frac{1}{2}\times\sqrt{2}\times1 = \frac{\sqrt{2}}{2}$.
4. 如图,在边长为4的正方形ABCD中,E是BC上一点,F是CD延长线上一点,连接AE,AF,AM平分∠EAF交CD于点M. 若BE = DF = 1,则DM的长度为( )

A. 2
B. $\sqrt{5}$
C. $\sqrt{6}$
D. $\frac{12}{5}$
A. 2
B. $\sqrt{5}$
C. $\sqrt{6}$
D. $\frac{12}{5}$
答案
D
5. 如图,正方形ABCD的边长为8,E是CD的中点,HG垂直平分AE且分别交AE,BC于点H,G,则BG = ____________.
答案
1
6.(2023·天津)如图,在边长为3的正方形ABCD的外侧,作等腰三角形ADE,EA = ED = $\frac{5}{2}$.
(1)△ADE的面积为________;
(2)若F为BE的中点,连接AF并延长,与CD相交于点G,则AG的长为________.
(1)△ADE的面积为________;
(2)若F为BE的中点,连接AF并延长,与CD相交于点G,则AG的长为________.
答案
3 $\sqrt{13}$
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