《2025年拔尖特训八年级数学上册人教版》

10. (1) 如图①，请探究∠ADB与∠A，∠B，∠ACB之间的数量关系，并给出证明.
(2) 如图②，DE平分∠ADB，CE平分∠ACB，∠A=24°，∠B=66°，求∠E的度数.

答案：

(1) $ \angle ADB = \angle A + \angle B + \angle ACB $。如图，连接 $ CD $ 并延长至点 $ O $。$ \therefore \angle ADO = \angle A + \angle ACO $，$ \angle BDO = \angle B + \angle BCO $。$ \because \angle ADB = \angle ADO + \angle BDO $，$ \therefore \angle ADB = \angle A + \angle ACO + \angle B + \angle BCO $。$ \therefore \angle ADB = \angle A + \angle B + \angle ACB $。
(2) 由
(1)，可得 $ \angle ADB = \angle A + \angle B + \angle ACB $，$ \angle ADE = \angle A + \angle E + \angle ACE $。$ \because DE $ 平分 $ \angle ADB $，$ CE $ 平分 $ \angle ACB $，$ \therefore \angle ADE = \frac{1}{2} \angle ADB $，$ \angle ACE = \frac{1}{2} \angle ACB $。$ \therefore \frac{1}{2} \angle ADB = \angle A + \angle E + \frac{1}{2} \angle ACB $，即 $ \angle ADB = 2 \angle A + 2 \angle E + \angle ACB $。$ \therefore \angle A + \angle B + \angle ACB = 2 \angle A + 2 \angle E + \angle ACB $。整理，得 $ \angle E = \frac{1}{2} (\angle B - \angle A) $。$ \because \angle A = 24^{\circ} $，$ \angle B = 66^{\circ} $，$ \therefore \angle E = \frac{1}{2} \times (66^{\circ} - 24^{\circ}) = 21^{\circ} $。

11. (2024·达州改编)如图，在△ABC中，AE₁，BE₁分别是内角∠CAB和外角∠CBD的三等分线，且∠E₁AD=1/3∠CAB，∠E₁BD=1/3∠CBD.
(1) 若∠C=60°，则∠E₁=______.
(2) 在△ABE₁中，AE₂，BE₂分别是内角∠E₁AB和外角∠E₁BD的三等分线，且∠E₂AD=1/3∠E₁AB，∠E₂BD=1/3∠E₁BD，…，以此规律继续作下去.若∠C=m°，试用含m，n的式子表示∠Eₙ的度数.

答案：

(1) $ 20^{\circ} $。解析：如图，依题意，得 $ \angle C = 60^{\circ} $，$ \angle 1 = \frac{1}{3} \angle CAB $，$ \angle 2 = \frac{1}{3} \angle CBD $，$ \angle CBD = \angle C + \angle CAB $，$ \therefore 3 \angle 2 = 60^{\circ} + 3 \angle 1 $，即 $ \angle 2 = 20^{\circ} + \angle 1 $。$ \because \angle 2 = \angle 1 + \angle E_1 $，$ \therefore 20^{\circ} + \angle 1 = \angle 1 + \angle E_1 $。$ \therefore \angle E_1 = 20^{\circ} $。
(2) 由题意，得 $ \angle E_1AD = \frac{1}{3} \angle CAB $，$ \angle E_1BD = \frac{1}{3} \angle CBD $，$ \therefore $ 设 $ \angle E_1AD = \alpha $，$ \angle E_1BD = \beta $，则 $ \angle CAB = 3 \alpha $，$ \angle CBD = 3 \beta $。由三角形的外角性质，得 $ \beta = \alpha + \angle E_1 $，$ 3 \beta = 3 \alpha + \angle C $，$ \therefore \angle E_1 = \frac{1}{3} \angle C $。同理，可得 $ \angle E_2 = \frac{1}{3} \angle E_1 = (\frac{1}{3})^2 \angle C $，$ \cdots $，$ \angle E_n = (\frac{1}{3})^n \angle C $。$ \therefore \angle E_n = (\frac{1}{3})^n \angle C $。