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11. ▶教材P125练习T3变式 [2025开封祥符区期末] 下列条件中，不能判断$△ ABC$是直角三角形的是（
A.$a:b:c=3:4:5$
B.$∠ A:∠ B:∠ C=3:4:5$
C.$∠ A+∠ B=∠ C$
D.$a:b:c=1:2:\sqrt{3}$
B
）A.$a:b:c=3:4:5$
B.$∠ A:∠ B:∠ C=3:4:5$
C.$∠ A+∠ B=∠ C$
D.$a:b:c=1:2:\sqrt{3}$
答案
选项A,$\because a:b:c = 3:4:5$,$\therefore$ 设$a = 3x$,$b = 4x$,$c = 5x$,则$(3x)^2 + (4x)^2 = (5x)^2$,$\therefore △ ABC$是直角三角形;选项B,$\because ∠ A:∠ B:∠ C = 3:4:5$,$\therefore$ 设$∠ A = 3x$,$∠ B = 4x$,$∠ C = 5x$,$\therefore 3x + 4x + 5x = 180°$,解得$x = 15°$,$\therefore 3x = 15° × 3 = 45°$,$4x = 15° × 4 = 60°$,$5x = 15° × 5 = 75°$,$\therefore △ ABC$是锐角三角形;选项C,$\because ∠ A + ∠ B = ∠ C$,$∠ A + ∠ B + ∠ C = 180°$,$\therefore ∠ C = 90°$,$\therefore △ ABC$是直角三角形;选项D,$\because a:b:c = 1:2:\sqrt{3}$,$\therefore$ 设$a = x$,$b = 2x$,$c = \sqrt{3}x$,则$x^2 + (\sqrt{3}x)^2 = (2x)^2$,$\therefore △ ABC$是直角三角形.
12. 如图,在$△ ABC$中,$AC=5$,$BC=12$,$AB=13$,将$△ ABC$沿$AD$折叠,使点$C$落在边$AB$上的点$E$处,求$DC$的长.
答案
解:$\because 5^2 + 12^2 = 13^2$,即$AC^2 + BC^2 = AB^2$,
$\therefore △ ABC$是直角三角形,$∠ C = 90°$.
根据题意,得$AE = AC = 5$,$DE = DC$,$∠ AED = ∠ C = 90°$,
$\therefore ∠ BED = 90°$.
设$DC = x$,则$DE = x$,$BD = 12 - x$.
在$\mathrm{Rt}△ BDE$中,由勾股定理,
得$x^2 + (13 - 5)^2 = (12 - x)^2$,解得$x = \frac{10}{3}$,$\therefore DC = \frac{10}{3}$.
$\therefore △ ABC$是直角三角形,$∠ C = 90°$.
根据题意,得$AE = AC = 5$,$DE = DC$,$∠ AED = ∠ C = 90°$,
$\therefore ∠ BED = 90°$.
设$DC = x$,则$DE = x$,$BD = 12 - x$.
在$\mathrm{Rt}△ BDE$中,由勾股定理,
得$x^2 + (13 - 5)^2 = (12 - x)^2$,解得$x = \frac{10}{3}$,$\therefore DC = \frac{10}{3}$.
13. 如图,在△ABC中,AC=4,BC=3,AB=5,AD为△ABC的角平分线,求△ABD的面积. 
答案
解:如图,过点$D$作$DE ⊥ AB$于点$E$.
$\because AC = 4$,$BC = 3$,$AB = 5$,$\therefore AC^2 + BC^2 = AB^2$,
$\therefore △ ABC$是直角三角形,且$∠ C = 90°$.
$\because AD$是$△ ABC$的角平分线,$DE ⊥ AB$于点$E$,
$\therefore CD = DE$(角平分线的性质).
在$\mathrm{Rt}△ ACD$和$\mathrm{Rt}△ AED$中,
$\because AD = AD$,$CD = ED$,$\therefore \mathrm{Rt}△ ACD ≌ \mathrm{Rt}△ AED(\mathrm{HL})$.
$\therefore AE = AC = 4$,$\therefore BE = AB - AE = 5 - 4 = 1$.
设$BD = x$,则$DE = CD = BC - BD = 3 - x$.
在$\mathrm{Rt}△ BDE$中,$BE^2 + DE^2 = BD^2$,
$\therefore 1^2 + (3 - x)^2 = x^2$,解得$x = \frac{5}{3}$.
$\therefore S_{△ ABD} = \frac{1}{2} BD · AC = \frac{1}{2} × \frac{5}{3} × 4 = \frac{10}{3}$.
14. ▶教材P142 复习题T13 变式|推理能力 △ABC 的三边分别为$a,b,c$,且$a^2 + b^2 + c^2 = 12a + 16b + 20c - 200$,请判断△ABC 的形状,并说明理由.
答案
解:$△ ABC$是直角三角形.理由如下:
$\because a^2 + b^2 + c^2 = 12a + 16b + 20c - 200$,
$\therefore a^2 - 12a + 36 + b^2 - 16b + 64 + c^2 - 20c + 100 = 0$,
$\therefore (a - 6)^2 + (b - 8)^2 + (c - 10)^2 = 0$(配成三数差的平方式),
$\therefore a - 6 = 0$,$b - 8 = 0$,$c - 10 = 0$,$\therefore a = 6$,$b = 8$,$c = 10$.
$\because 6^2 + 8^2 = 10^2$,$\therefore △ ABC$是直角三角形.
$\because a^2 + b^2 + c^2 = 12a + 16b + 20c - 200$,
$\therefore a^2 - 12a + 36 + b^2 - 16b + 64 + c^2 - 20c + 100 = 0$,
$\therefore (a - 6)^2 + (b - 8)^2 + (c - 10)^2 = 0$(配成三数差的平方式),
$\therefore a - 6 = 0$,$b - 8 = 0$,$c - 10 = 0$,$\therefore a = 6$,$b = 8$,$c = 10$.
$\because 6^2 + 8^2 = 10^2$,$\therefore △ ABC$是直角三角形.
15. 推理能力 [2025哈尔滨南岗区期末] 新定义:如图，点M,N(点M在点N的左侧)把线段AB分割成AM,MN,NB三条线段，若以线段AM,MN,NB为边的三角形是直角三角形，则称点M,N是线段AB的勾股分割点.
(1)若$AM=1.5$,$MN=2.5$,$BN=2$,则点M,N是线段AB的勾股分割点吗？请说明理由.
(2)若点M,N是线段AB的勾股分割点,且AM为直角边,$AB=30$,$AM=5$,求BN的长.
(1)若$AM=1.5$,$MN=2.5$,$BN=2$,则点M,N是线段AB的勾股分割点吗？请说明理由.
(2)若点M,N是线段AB的勾股分割点,且AM为直角边,$AB=30$,$AM=5$,求BN的长.
答案
解:(1)是.理由如下:
$\because AM^2 + BN^2 = 1.5^2 + 2^2 = 6.25$,$MN^2 = 2.5^2 = 6.25$,
$\therefore AM^2 + NB^2 = MN^2$,
$\therefore$ 以$AM$,$MN$,$NB$为边的三角形是直角三角形,
$\therefore$ 点$M$,$N$是线段$AB$的勾股分割点.
(2)设$BN = x$,则$MN = 30 - AM - BN = 25 - x$.
①当$MN$为斜边时,得$MN^2 = AM^2 + NB^2$,
即$(25 - x)^2 = x^2 + 25$,解得$x = 12$.
②当$BN$为斜边时,得$BN^2 = AM^2 + MN^2$,
即$x^2 = 25 + (25 - x)^2$,解得$x = 13$.
综上所述,$BN$的长为12或13.
$\because AM^2 + BN^2 = 1.5^2 + 2^2 = 6.25$,$MN^2 = 2.5^2 = 6.25$,
$\therefore AM^2 + NB^2 = MN^2$,
$\therefore$ 以$AM$,$MN$,$NB$为边的三角形是直角三角形,
$\therefore$ 点$M$,$N$是线段$AB$的勾股分割点.
(2)设$BN = x$,则$MN = 30 - AM - BN = 25 - x$.
①当$MN$为斜边时,得$MN^2 = AM^2 + NB^2$,
即$(25 - x)^2 = x^2 + 25$,解得$x = 12$.
②当$BN$为斜边时,得$BN^2 = AM^2 + MN^2$,
即$x^2 = 25 + (25 - x)^2$,解得$x = 13$.
综上所述,$BN$的长为12或13.
