1[2026江苏南通调研,中]一个等腰三角形,其中两条边长度的比是2:5,其中一条边长度是10 cm,这个等腰三角形的周长最大可以是
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A.18 cm
B.24 cm
C.45 cm
D.60 cm

1. D 【解析】$\because$ 等腰三角形两边长度之比为 $2:5,\therefore$ 设这两边长为 $2x,5x(x＞0)$. 若腰长为 $2x$,底边长为 $5x$,此时三边长为 $2x,2x,5x$. $\because 2x+2x＜5x,\therefore$ 无法构成三角形,三角形不存在. 若腰长为 $5x$,底边长为 $2x$,此时三边长为 $5x,5x,2x$. $\because 2x+5x＞5x,\therefore$ 可以构成三角形. 当 $2x=10$ 时,$x=5$,则 $5x=25,\therefore$ 此时周长为 $25+25+10=60(\mathrm{cm})$. 当 $5x=10$ 时,$x=2$,则 $2x=4,\therefore$ 此时周长为 $10+10+4=24(\mathrm{cm})$.$\therefore$ 这个等腰三角形的周长最大可以是 $60\ \mathrm{cm}$,故选 D.

2[中]如图,在$△ ABC$中,$AB=AC$,E是BC边上一点,将$△ ABE$沿AE翻折,使点B落到点D的位置,AD边与BC边交于点F,如果$AE=EF=DF$,那么$∠ BAE$的度数为.

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2. $(\frac{360}{7})°$ 【解析】$\because AB=AC,\therefore ∠ B = ∠ C$. 令$∠ B = ∠ C = α$,由折叠的性质可得$∠ D = ∠ B = α. \because EF = DF,\therefore ∠ D = ∠ FED = α,\therefore ∠ AFE = ∠ D + ∠ FED = 2α. \because AE = EF,\therefore ∠ EAF = ∠ AFE = 2α$. 由翻折可知$∠ BAE = ∠ EAF = 2α$,$\therefore ∠ AEF = ∠ B + ∠ BAE = 3α$. 在 $△ AEF$ 中,$∠ AFE + ∠ FEA + ∠ EAF = 180°$,即 $2α+3α+2α=180°$,解得 $α = (\frac{180}{7})°$,$\therefore ∠ BAE = 2α = (\frac{360}{7})°$. 故答案为 $(\frac{360}{7})°$.

3[中]如图,在$△ ABC$中,点$D$为$BC$边上一点,$BD=BA$. $EF$垂直平分$AC$,交$AC$于点$E$,交$BC$于点$F$,连接$AF$,$AD$. 当$∠ B=30°$,$∠ BAF=90°$时,$∠ DAC$的度数为$°$.

3. 45 【解析】$\because BA = BD, ∠ B = 30°, \therefore ∠ BAD = ∠ BDA = \frac{1}{2}(180°-∠ B) = 75°. \because ∠ BAF = 90°, \therefore ∠ AFB = 90° - ∠ B = 60°, \therefore ∠ C + ∠ CAF = 60°. \because EF$ 垂直平分 $AC, \therefore FA = FC, \therefore ∠ C = ∠ CAF = 30°, \therefore ∠ DAC = ∠ ADB - ∠ C = 45°$. 故答案为 45.

4[中]如图,等腰△ABC底边BC的长为6 cm,面积是18 cm²,腰AB的垂直平分线EF交AC于点F,交AB于点E,若D为BC边上的中点,M为线段EF上一个动点,则△BDM周长的最小值为 cm.

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4. 9 【解析】连接 $AM, AD. \because △ ABC$ 是等腰三角形,点 $D$ 是 $BC$ 边的中点, $\therefore BD = \frac{1}{2}BC = 3\ \mathrm{cm}, AD ⊥ BC, S_{△ ABC} = \frac{1}{2}BC · AD = \frac{1}{2}×6× AD = 18\ \mathrm{cm}^2, \therefore AD = 6\ \mathrm{cm}. \because EF$ 是线段 $AB$ 的垂直平分线, $\therefore AM = BM, \therefore △ BDM$ 的周长为 $BM+MD+BD = AM+DM+3$. 又$\because AM+DM \ge AD, \therefore$ 当 $A,M,D$ 三点共线时, $△ BDM$ 周长有最小值, 为 $AD+3=6+3=9(\mathrm{cm})$. 故答案为 9.

5[2026河南信阳质检,较难]如图,在$△ ABC$中,$AB=AC$,$∠ A=40°$,$D$是边$AB$上的动点,连接$CD$,将$△ ADC$沿直线$CD$翻折得到$△ A'DC$,直线$AB$与直线$A'C$交于点$E$.若$△ A'DE$是等腰三角形,则$∠ ACD$的度数为.

5. $15°或30°$ 【解析】设 $∠ ACD = α. \because △ ADC$ 沿直线 $CD$ 翻折得到 $△ A'DC, \therefore ∠ A = ∠ A' = 40°, ∠ ACD = ∠ A'CD = α, ∠ ADC = ∠ A'DC = 140°-α, \therefore ∠ AEA' = 2α+40°, ∠ A'DE = 100°-2α$. 如图, 当 $A'E = A'D$ 时, $∠ A'DE = ∠ AEA', \therefore 100°-2α=2α+40°, \therefore α=15°$. 如图, 当 $A'E = DE$ 时, $∠ A'DE = ∠ A', \therefore 100°-2α=40°, \therefore α=30°$. 综上, $∠ ACD$ 的度数为 $15°$或 $30°$. 故答案为 $15°$或 $30°$.

6[中]如图,在$△ ABC$中,$AB=AC$,点$O$在高$AE$上,且$OA=OB$,连接$BO$并延长交$AC$于点$D$.
(1)求证:$∠ BAC=2∠ ABD$.
(2)若$△ BCD$是等腰三角形,求$∠ BAC$的度数.

6. (1)【证明】$\because AB = AC, AE ⊥ BC, \therefore ∠ BAC = 2∠ BAE. \because OA = OB, \therefore ∠ ABD = ∠ BAE, \therefore ∠ BAC = 2∠ ABD$.
(2)【解】①当 $BD = BC$ 时, $∠ C = ∠ BDC$. $\because ∠ ABD = ∠ BAE = ∠ CAE, \therefore ∠ BDC = ∠ ABD + ∠ BAC = 3∠ ABD$. 设 $∠ ABD = α$, 则 $∠ BAC = 2α, ∠ BDC = ∠ C = ∠ ABC = 3α, \therefore 2α+3α+3α=180°, \therefore α=22.5°, \therefore 2α=45°, \therefore ∠ BAC=45°$.
②当 $BC = CD$ 时, $∠ CBD = ∠ CDB, \therefore ∠ CBD = ∠ CDB = 3∠ ABD$. 设 $∠ ABD = β$, 则 $∠ BAC = 2β, ∠ CBD = ∠ CDB = 3β, \therefore ∠ ABC = ∠ C = 4β$. $\because ∠ ABC + ∠ C + ∠ BAC = 180°, \therefore 4β+4β+2β=180°, \therefore β=18°, \therefore 2β=36°, \therefore ∠ BAC=36°$.
综上所述,$∠ BAC$ 的度数为 $45°$或 $36°$.

7[中]如图,在等腰△ABC中,AB=AC,点D在BC上,且AD=AE.
(1)若∠BAC=90°,∠BAD=30°,求∠EDC的度数.
(2)若∠BAC=α(α＞30°),∠BAD=30°,求∠EDC的度数.
(3)猜想∠EDC与∠BAD的数量关系.(不必证明)

7. 【解】(1) $\because ∠ BAC = 90°, AB = AC, \therefore ∠ B = ∠ C = \frac{1}{2}(180° - ∠ BAC) = 45°, \therefore ∠ ADC = ∠ B + ∠ BAD = 45° + 30° = 75°. \because ∠ DAC = ∠ BAC - ∠ BAD = 90° - 30° = 60°, AD = AE, \therefore ∠ ADE = ∠ AED = \frac{1}{2}(180° - ∠ DAC) = 60°, \therefore ∠ EDC = ∠ ADC - ∠ ADE = 75°-60°=15°$.
(2) $\because ∠ BAC = α, AB = AC, \therefore ∠ B = ∠ C = \frac{1}{2}(180° - ∠ BAC) = 90° - \frac{1}{2}α, \therefore ∠ ADC = ∠ B + ∠ BAD = 90° - \frac{1}{2}α + 30° = 120° - \frac{1}{2}α$.
$\because AD = AE, ∠ DAC = ∠ BAC - ∠ BAD = α - 30°, \therefore ∠ ADE = ∠ AED = \frac{1}{2}(180° - ∠ DAC) = 105° - \frac{1}{2}α, \therefore ∠ EDC = ∠ ADC - ∠ ADE = (120° - \frac{1}{2}α) - (105° - \frac{1}{2}α) = 15°$, 即 $∠ EDC$ 的度数是 $15°$.
(3) $∠ EDC$ 与 $∠ BAD$ 的数量关系是 $∠ EDC = \frac{1}{2}∠ BAD$.

8 思想方法 数形结合 [2025江苏扬州质检,较难]某数学兴趣小组开展了一次活动,过程如下:如图,已知$∠ BAC=θ(0°＜θ＜90°)$.现把等长的小棒依次摆放在射线$AB$,$AC$之间,并使小棒两端分别落在射线$AB$,$AC$上.$A_1A_2$为第一根小棒,且$A_1A_2=AA_1$,若只能摆放4根小棒,则$θ$的范围为.

8. $18° \le θ ＜ 22.5°$ 【解析】$\because AA_1 = A_1A_2, \therefore ∠ AA_2A_1 = ∠ A, \therefore ∠ A_2A_1A_3 = ∠ AA_2A_1 + ∠ A = 2∠ BAC = 2θ. \because A_1A_2 = A_2A_3, \therefore ∠ A_2A_1A_3 = ∠ A_2A_3A_1 = 2θ, \therefore ∠ A_3A_2A_4 = ∠ A + ∠ A_2A_3A_1 = θ + 2θ = 3θ. \because A_2A_3 = A_3A_4, \therefore ∠ A_3A_2A_4 = ∠ A_3A_4A = 3θ, \therefore ∠ A_4A_3C = ∠ A_3A_4A + ∠ A = 4θ$. 同理, $∠ A_5A_4B = 5θ. \because$ 只能摆放 4 根小棒, $\therefore 4θ＜90°$ 且 $5θ\ge90°$, 解得 $18° \le θ ＜ 22.5°$, 故答案为 $18° \le θ ＜ 22.5°$.