《2025年亮点给力大试卷八年级数学上册苏科版》

22. (4分)新素养 推理能力 如图,在$△ABC$中,$AB = AC$,D是BA延长线上一点,E是AC的中点,连接DE并延长,交BC于点M,$∠DAC$的平分线交DM于点F.求证：$AF = CM$.

答案： 因为AB = AC,所以∠B = ∠C,所以∠DAC = ∠B + ∠C = 2∠C.因为AF平分∠DAC,所以∠DAC = 2∠EAF,所以∠EAF = ∠C.因为E是AC的中点,所以AE = CE.在△AEF和△CEM中,$\left\{\begin{array}{l} ∠EAF = ∠C\\ AE = CE\\ ∠AEF = ∠CEM\end{array}\right.$，所以△AEF≌△CEM(ASA),所以AF = CM.

23. (6分)(2025·江苏泰州期末)问题：如图,在$△ABD$中,$BA = BD$,在BD的延长线上取点E,C,作$△AEC$,使$EA = EC$.若$∠BAE = 90°$,$∠B = 45°$,求$∠DAC$的度数.
答案：$∠DAC = 45°$.
思考：
(1) 如果把以上“问题”中的条件“$∠B = 45°$”去掉,其余条件不变,那么$∠DAC$的度数会改变吗？说明理由；
(2) 如果把以上“问题”中的条件“$∠B = 45°$”去掉,再将“$∠BAE = 90°$”改为“$∠BAE = n°$”,其余条件不变,求$∠DAC$的度数.

答案：
(1)∠DAC的度数不会改变.理由如下:因为BA = BD,所以∠ADB = ∠DAB = $\frac{1}{2}$(180° - ∠B).因为EA = EC,所以∠CAE = ∠C,所以∠AEB = ∠CAE + ∠C = 2∠C,所以∠C = $\frac{1}{2}$∠AEB.因为∠BAE = 90°,所以∠AEB = 90° - ∠B,所以∠C = $\frac{1}{2}$(90° - ∠B),所以∠DAC = ∠ADB - ∠C = 45°.
(2)因为BA = BD,所以∠ADB = ∠DAB = $\frac{1}{2}$(180° - ∠B).因为EA = EC,所以∠CAE = ∠C,所以∠AEB = ∠CAE + ∠C = 2∠C,所以∠C = $\frac{1}{2}$∠AEB = $\frac{1}{2}$(180° - ∠BAE - ∠B),所以∠DAC = ∠ADB - ∠C = $\frac{1}{2}$∠BAE.因为∠BAE = n°,所以∠DAC = ($\frac{n}{2}$)°.

24. (6分)已知$△ABC$.
(1) 请在图中用无刻度的直尺和圆规作图：作$∠BAC$的平分线和BC的垂直平分线,它们的交点为D；(保留作图痕迹,不写作法)
(2) 若$AB = 15$,$AC = 9$,过点D作$DE⊥AB$于点E,则BE的长为______.(如需画草图,请使用备用图)

答案：

(1)图略.
(2)3　解析:如图,连接DB,DC,过点D作DH⊥AC,交AC的延长线于点H,则∠DHA = 90°.因为DE⊥AB,所以∠DEA = ∠DEB = 90°,所以∠DEA = ∠DHA.因为AD平分∠BAC,所以∠DAE = ∠DAH.在△DAE和△DAH中,$\left\{\begin{array}{l} ∠DEA = ∠DHA\\ ∠DAE = ∠DAH\\ AD = AD\end{array}\right.$，所以△DAE≌△DAH(AAS),所以AE = AH,DE = DH.因为点D在BC的垂直平分线上,所以DB = DC.在Rt△DBE和Rt△DCH中,$\left\{\begin{array}{l} DB = DC\\ DE = DH\end{array}\right.$，所以Rt△DBE≌Rt△DCH(HL),所以BE = CH.设BE = CH = x.因为AB = 15,AC = 9,所以AE = AB - BE = 15 - x,AH = AC + CH = 9 + x,所以15 - x = 9 + x,解得x = 3,即BE的长为3.