《2025年多维导学案八年级数学上册人教版》

5. 如图,在$\triangle ABC$中,$∠C=3∠B$,AD平分$∠BAC$交BC于D.
(1)如图1,若$AE⊥BC$于点E,$∠C=75^{\circ }$,求$∠DAE$的度数;
(2)如图2,若$DF⊥AD$交AB于点F,求证:$BF=DF.$

答案： 解:
(1)
∵ $∠C = 3∠B$, $∠C = 75^{\circ}$,
∴ $∠B = 25^{\circ}$.
∴ $∠BAC = 180^{\circ} - ∠B - ∠C = 80^{\circ}$.
∵ AD 平分 $∠BAC$,
∴ $∠BAD = \frac{1}{2}∠BAC = 40^{\circ}$.
∴ $∠ADE = ∠BAD + ∠B = 65^{\circ}$.
∵ $AE ⊥ BC$,
∴ $∠AED = 90^{\circ}$.
∴ $∠DAE = 90^{\circ} - ∠ADE = 90^{\circ} - 65^{\circ} = 25^{\circ}$.
(2) 证明: 设 $∠B = \alpha$, 则 $∠C = 3\alpha$.
∴ $∠BAC = 180^{\circ} - ∠B - ∠C = 180^{\circ} - 4\alpha$.
∵ AD 平分 $∠BAC$,
∴ $∠BAD = \frac{1}{2}∠BAC = 90^{\circ} - 2\alpha$.
∵ $DF ⊥ AD$,
∴ $∠ADF = 90^{\circ}$.
∴ $∠AFD = 90^{\circ} - ∠BAD = 2\alpha$.
∵ $∠AFD = ∠B + ∠BDF$,
∴ $∠BDF = \alpha = ∠B$.
∴ $BF = DF$.

6. 如图,在$\triangle ABC$中,$AB=AC=2,∠B=∠C=40^{\circ }$,点D在线段BC上运动(点D不与点B,C重合),连接AD,作$∠ADE=40^{\circ }$,DE交线段AC于点E.
(1)当$∠BDA=115^{\circ }$时,$∠EDC=$25$^{\circ },∠DEC=$115$^{\circ }$;点D从点B向点C运动时,$∠BDA$逐渐变小;(填“大”或“小”)
(2)当DC等于多少时,$\triangle ABD\cong \triangle DCE$,请说明理由;解:当$DC = 2$时,$△ABD ≌ △DCE$.理由如下: ∵ $∠B = ∠C = 40^{\circ}$,∴ $∠DEC + ∠EDC = 140^{\circ}$.又 ∵ $∠ADE = 40^{\circ}$,∴ $∠ADB + ∠EDC = 140^{\circ}$.∴ $∠ADB = ∠DEC$.又 ∵ $AB = DC = 2$,∴ $△ABD ≌ △DCE$.
(3)在点D的运动过程中,$\triangle ADE$的形状可以是等腰三角形吗? 若可以,请求出$∠BDA$的度数;若不可以,请说明理由.解:当$∠BDA$的度数为$110^{\circ}$或$80^{\circ}$时,$△ADE$的形状是等腰三角形.理由如下:∵ $∠C = ∠ADE = 40^{\circ}$,$∠AED = ∠C + ∠EDC$,∴ $∠AED ＞ ∠ADE$.∴ 当$△ADE$是等腰三角形时,只存在$AD = ED$或$AE = DE$两种情况.① 当$AD = ED$时,$∠DAE = ∠DEA$.∵ $∠ADE = 40^{\circ}$,∴ $∠AED = \frac{180^{\circ} - ∠ADE}{2} = 70^{\circ}$.∴ $∠EDC = ∠AED - ∠C = 30^{\circ}$.∴ $∠ADC = ∠ADE + ∠EDC = 70^{\circ}$.∴ $∠BDA = 180^{\circ} - ∠ADC = 110^{\circ}$;② 当$AE = DE$时,$∠EAD = ∠EDA = 40^{\circ}$,∴ $∠AED = 180^{\circ} - ∠EAD - ∠EDA = 100^{\circ}$.∴ $∠EDC = ∠AED - ∠C = 60^{\circ}$.∴ $∠ADC = ∠ADE + ∠EDC = 100^{\circ}$.∴ $∠BDA = 180^{\circ} - ∠ADC = 80^{\circ}$.综上所述,当$∠BDA$的度数为$110^{\circ}$或$80^{\circ}$时,$△ADE$的形状是等腰三角形.

答案： 解:
(1) 25 115 小
(2) 当 $DC = 2$ 时, $△ABD ≌ △DCE$.
理由如下:
∵ $∠B = ∠C = 40^{\circ}$,
∴ $∠DEC + ∠EDC = 140^{\circ}$.
又
∵ $∠ADE = 40^{\circ}$,
∴ $∠ADB + ∠EDC = 140^{\circ}$.
∴ $∠ADB = ∠DEC$.
又
∵ $AB = DC = 2$,
∴ $△ABD ≌ △DCE$.
(3) 当 $∠BDA$ 的度数为 $110^{\circ}$ 或 $80^{\circ}$ 时, $△ADE$ 的形状是等腰三角形. 理由如下:
∵ $∠C = ∠ADE = 40^{\circ}$,
$∠AED = ∠C + ∠EDC$,
∴ $∠AED ＞ ∠ADE$.
∴ 当 $△ADE$ 是等腰三角形时, 只存在 $AD = ED$ 或 $AE = DE$ 两种情况.
① 当 $AD = ED$ 时, $∠DAE = ∠DEA$.
∵ $∠ADE = 40^{\circ}$,
∴ $∠AED = \frac{180^{\circ} - ∠ADE}{2} = 70^{\circ}$.
∴ $∠EDC = ∠AED - ∠C = 30^{\circ}$.
∴ $∠ADC = ∠ADE + ∠EDC = 70^{\circ}$.
∴ $∠BDA = 180^{\circ} - ∠ADC = 110^{\circ}$;
② 当 $AE = DE$ 时, $∠EAD = ∠EDA = 40^{\circ}$,
∴ $∠AED = 180^{\circ} - ∠EAD - ∠EDA = 100^{\circ}$.
∴ $∠EDC = ∠AED - ∠C = 60^{\circ}$.
∴ $∠ADC = ∠ADE + ∠EDC = 100^{\circ}$.
∴ $∠ADB = 180^{\circ} - ∠ADC = 80^{\circ}$.
综上所述, 当 $∠BDA$ 的度数为 $110^{\circ}$ 或 $80^{\circ}$ 时, $△ADE$ 的形状是等腰三角形.

7. 如图,$\triangle ABC,\triangle CDE$都是等边三角形,AD,BE相交于点O,点M,N分别是线段AD,BE的中点.
(1)求证:$AD=BE;$
(2)求$∠DOE$的度数;
(3)求证:$\triangle MNC$是等边三角形.

答案： 解:
(1) 证明:
∵ $△ABC$, $△CDE$ 都是等边三角形,
∴ $AC = BC$, $CD = CE$, $∠ACB = ∠DCE = 60^{\circ}$.
∴ $∠ACB + ∠BCD = ∠DCE + ∠BCD$.
∴ $∠ACD = ∠BCE$.
在 $△ACD$ 和 $△BCE$ 中,
$\begin{cases}AC = BC, \\∠ACD = ∠BCE, \\CD = CE,\end{cases}$
∴ $△ACD ≌ △BCE(SAS)$.
∴ $AD = BE$.
(2) 由
(1), 得 $△ACD ≌ △BCE$,
∴ $∠ADC = ∠BEC$.
∵ $△DCE$ 是等边三角形,
∴ $∠CED = ∠CDE = 60^{\circ}$.
∴ $∠ADE + ∠BED = ∠ADC + ∠CDE + ∠BED$
$= ∠ADC + 60^{\circ} + ∠BED$
$= ∠BEC + ∠BED + 60^{\circ}$
$= ∠CED + 60^{\circ}$
$= 60^{\circ} + 60^{\circ}$
$= 120^{\circ}$.
∴ $∠DOE = 180^{\circ} - (∠ADE + ∠BED) = 60^{\circ}$.
∴ $∠DOE$ 的度数是 $60^{\circ}$.
(3) 证明: 由
(1), 得 $△ACD ≌ △BCE$,
∴ $∠CAD = ∠CBE$, $AD = BE$, $AC = BC$.
又
∵ 点 M, N 分别是线段 AD, BE 的中点,
∴ $AM = \frac{1}{2}AD$, $BN = \frac{1}{2}BE$.
∴ $AM = BN$.
在 $△ACM$ 和 $△BCN$ 中,
$\begin{cases}AC = BC, \\∠CAM = ∠CBN, \\AM = BN,\end{cases}$
∴ $△ACM ≌ △BCN(SAS)$.
∴ $CM = CN$, $∠ACM = ∠BCN$.
又 $∠ACB = 60^{\circ}$,
∴ $∠ACM + ∠MCB = 60^{\circ}$.
∴ $∠BCN + ∠MCB = 60^{\circ}$.
∴ $∠MCN = 60^{\circ}$.
∴ $△MNC$ 是等边三角形.