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1.「2026陕西西安理工大附中月考，★☆」如图，在平面直角坐标系xOy中，点A的坐标为(10,8)，过点A作AB⊥x轴于点B，AC⊥y轴于点C，点D在AB上。将△CAD沿直线CD翻折，点A恰好落在x轴上的点E处，则点D的坐标为（
A.(10,4)
B.(10,3)
C.(10,2.5)
D.(10,2)
B
）A.(10,4)
B.(10,3)
C.(10,2.5)
D.(10,2)
答案
1.B 设 $DB = m.$ 由题意可得 $OB = CA = 10, OC = AB = 8, CE = CA = 10, DE = DA = 8 - m.$ 在 $\mathrm{Rt}△ COE$ 中, $OE = \sqrt{CE^2 - OC^2} = \sqrt{10^2 - 8^2} = 6,\therefore EB = 10 - 6 = 4.$ 在 $\mathrm{Rt}△ DBE$ 中, $∠ DBE = 90°,$ $\therefore DE^2 = DB^2 + EB^2,$ 即 $(8 - m)^2 = m^2 + 4^2,$ 解得 $m = 3,\therefore$ 点 $D$ 的坐标是 $(10,3).$ 故选 B.
2.「2026陕西西安月考改编，★★☆」如图,在矩形$ABCD$中,$AB=4$,$BC=3$,$P$为边$BC$上一点.将$△ CDP$沿$DP$所在直线折叠,点$C$落在点$E$处,$DE$,$PE$分别交$AB$于点$F$,$G$,已知$GE=GB$,则$BF$的长为（
A.$\dfrac{17}{5}$
B.$\dfrac{3}{5}$
C.$\dfrac{12}{5}$
D.$5$
C
）A.$\dfrac{17}{5}$
B.$\dfrac{3}{5}$
C.$\dfrac{12}{5}$
D.$5$
答案
2.C $\because$ 四边形 $ABCD$ 是矩形，$\therefore AB = CD = 4, AD = BC = 3, ∠ A = ∠ B = ∠ C = 90°.$
由折叠得 $DE = CD = 4, EP = PC, ∠ E = ∠ C = 90°.$
在 $△ GEF$ 和 $△ GBP$ 中, $\begin{cases}∠ EGF = ∠ BGP,\\GE = GB,\\∠ E = ∠ B,\end{cases}$
$\therefore △ GEF ≌ △ GBP(\mathrm{ASA}),\therefore EF = BP, GF = GP,$
$\because BF = BG + FG, EP = EG + PG,$
$\therefore BF = EP = CP.$
设 $BF = EP = CP = x,$ 则 $AF = 4 - x, EF = BP = 3 - x,$
$\therefore DF = DE - EF = 4 - (3 - x) = 1 + x,$
在 $\mathrm{Rt}△ ADF$ 中, 由勾股定理可得 $AF^2 + AD^2 = DF^2,$
$\therefore (4 - x)^2 + 3^2 = (1 + x)^2,$ 解得 $x = \dfrac{12}{5},\therefore BF = \dfrac{12}{5}.$
故选 C.
由折叠得 $DE = CD = 4, EP = PC, ∠ E = ∠ C = 90°.$
在 $△ GEF$ 和 $△ GBP$ 中, $\begin{cases}∠ EGF = ∠ BGP,\\GE = GB,\\∠ E = ∠ B,\end{cases}$
$\therefore △ GEF ≌ △ GBP(\mathrm{ASA}),\therefore EF = BP, GF = GP,$
$\because BF = BG + FG, EP = EG + PG,$
$\therefore BF = EP = CP.$
设 $BF = EP = CP = x,$ 则 $AF = 4 - x, EF = BP = 3 - x,$
$\therefore DF = DE - EF = 4 - (3 - x) = 1 + x,$
在 $\mathrm{Rt}△ ADF$ 中, 由勾股定理可得 $AF^2 + AD^2 = DF^2,$
$\therefore (4 - x)^2 + 3^2 = (1 + x)^2,$ 解得 $x = \dfrac{12}{5},\therefore BF = \dfrac{12}{5}.$
故选 C.
3.「2026山西临汾期末，★☆」如图，在矩形ABCD中，点E在边CD上，将矩形ABCD沿AE所在直线折叠，点D恰好落在边BC上的点F处.若AB=8，DE=5，则折痕AE的长为
$5\sqrt{5}$
.答案
3.答案 $5\sqrt{5}$
解析 $\because$ 四边形 $ABCD$ 是矩形, $\therefore AB = CD = 8, BC = AD,$
$∠ B = ∠ D = ∠ C = 90°, \therefore CE = CD - DE = 8 - 5 = 3,$ 由折叠得 $FE = DE = 5, AF = AD, \therefore$ 在 $\mathrm{Rt}△ ECF$ 中, $CF = \sqrt{EF^2 - CE^2} = \sqrt{5^2 - 3^2} = 4,$ 设 $AD = x,$ 则 $BC = AF = x,$ 则 $BF = x - 4,$ 在 $\mathrm{Rt}△ ABF$ 中, 由勾股定理得 $8^2 + (x - 4)^2 = x^2,$ 解得 $x = 10,$
$\therefore AD = 10, \therefore AE = \sqrt{AD^2 + DE^2} = \sqrt{10^2 + 5^2} = 5\sqrt{5}.$
故答案为 $5\sqrt{5}.$
解析 $\because$ 四边形 $ABCD$ 是矩形, $\therefore AB = CD = 8, BC = AD,$
$∠ B = ∠ D = ∠ C = 90°, \therefore CE = CD - DE = 8 - 5 = 3,$ 由折叠得 $FE = DE = 5, AF = AD, \therefore$ 在 $\mathrm{Rt}△ ECF$ 中, $CF = \sqrt{EF^2 - CE^2} = \sqrt{5^2 - 3^2} = 4,$ 设 $AD = x,$ 则 $BC = AF = x,$ 则 $BF = x - 4,$ 在 $\mathrm{Rt}△ ABF$ 中, 由勾股定理得 $8^2 + (x - 4)^2 = x^2,$ 解得 $x = 10,$
$\therefore AD = 10, \therefore AE = \sqrt{AD^2 + DE^2} = \sqrt{10^2 + 5^2} = 5\sqrt{5}.$
故答案为 $5\sqrt{5}.$
4. 学科特色 方程思想 「2026 山东东营月考改编，★☆」如图，将长方形ABCD沿着对角线BD折叠，使点C落在C'处，BC'交AD于点E.
(1)若∠DBC=25°,求∠ADC'的度数.
(2)若AB=4,AD=8,求△BDE的面积.
(1)若∠DBC=25°,求∠ADC'的度数.
(2)若AB=4,AD=8,求△BDE的面积.
答案
4.解析 (1)$\because$ 四边形 $ABCD$ 是矩形，
$\therefore ∠ C = 90°, AD // BC,$
$\therefore ∠ DBC = ∠ ADB = 25°,$
由翻折可知 $∠ BDC = ∠ BDC' = 90° - 25° = 65°,$
$\therefore ∠ ADC' = ∠ BDC' - ∠ ADB = 65° - 25° = 40°.$
(2)由折叠可知 $∠ CBD = ∠ EBD,$
$\because AD // BC, \therefore ∠ CBD = ∠ EDB,$
$\therefore ∠ EBD = ∠ EDB, \therefore BE = DE,$
$\therefore △ BDE$ 是等腰三角形，
设 $DE = x,$ 则 $BE = x, AE = 8 - x,$
在 $\mathrm{Rt}△ ABE$ 中, 由勾股定理得 $AB^2 + AE^2 = BE^2,$
即 $4^2 + (8 - x)^2 = x^2,$ 解得 $x = 5,$
$\therefore S_{△ BDE} = \dfrac{1}{2}DE · AB = \dfrac{1}{2} × 5 × 4 = 10.$
$\therefore ∠ C = 90°, AD // BC,$
$\therefore ∠ DBC = ∠ ADB = 25°,$
由翻折可知 $∠ BDC = ∠ BDC' = 90° - 25° = 65°,$
$\therefore ∠ ADC' = ∠ BDC' - ∠ ADB = 65° - 25° = 40°.$
(2)由折叠可知 $∠ CBD = ∠ EBD,$
$\because AD // BC, \therefore ∠ CBD = ∠ EDB,$
$\therefore ∠ EBD = ∠ EDB, \therefore BE = DE,$
$\therefore △ BDE$ 是等腰三角形，
设 $DE = x,$ 则 $BE = x, AE = 8 - x,$
在 $\mathrm{Rt}△ ABE$ 中, 由勾股定理得 $AB^2 + AE^2 = BE^2,$
即 $4^2 + (8 - x)^2 = x^2,$ 解得 $x = 5,$
$\therefore S_{△ BDE} = \dfrac{1}{2}DE · AB = \dfrac{1}{2} × 5 × 4 = 10.$
