《2025年亮点给力大试卷八年级数学上册苏科版》

26. (6分)如图,在$Rt\triangle ABC$中,$∠BAC= 90^{\circ }$,D是BC的中点,作$DE⊥AC$,垂足为E.
(1) 求证:E是AC的中点;
(2) 将直角边AC沿点A,D确定的直线翻折,得到对应线段$AC'$.当$AC'⊥BC$时,判断$\triangle ABD$的形状,并说明理由.

答案：
(1)连接AD.因为∠BAC=90°,D是BC的中点,所以AD=BD=CD=$\frac {1}{2}BC$.因为DE⊥AC,所以E是AC的中点.
(2)△ABD为等边三角形.理由如下:设AC'交BC于点F.因为AC'⊥BC,所以∠AFC=90°.因为AD=CD,所以∠CAD=∠C.由折叠的性质,得∠C'AD=∠CAD=∠C.因为∠C'AD+∠CAD+∠C=90°,所以3∠C=90°,所以∠C=30°.因为∠BAC=90°,所以∠B=90°-∠C=60°.因为AD=BD,所以△ABD为等边三角形. 

27. (8分)新趋势 推导探究
(1) 如图,在$\triangle ABC$中,$∠BAC= 90^{\circ },AB= AC$,点D在CB的延长线上,且$DB= AB$,点E在BC的延长线上,且$EC= AC$,求$∠DAE$的度数;
(2) 若把(1)中“$AB= AC$”的条件去掉,其余条件不变,则$∠DAE$的度数会改变吗? 请说明理由;
(3) 若把(1)中“$∠BAC= 90^{\circ }$”的条件改为“$∠BAC＜90^{\circ }$”,其余条件不变,则$∠DAE与∠BAC$之间的数量关系为______.

(1)因为∠BAC = 90°,AB = AC,所以∠ABC = ∠ACB = $\frac{1}{2}(180^{\circ} - ∠BAC) = 45^{\circ}$.因为DB = AB,EC = AC,所以∠BAD = ∠D,∠CAE = ∠E,所以∠ABC = ∠BAD + ∠D = 2∠BAD,∠ACB = ∠CAE + ∠E = 2∠CAE,所以∠BAD = $\frac{1}{2}∠ABC = 22.5^{\circ}$,∠CAE = $\frac{1}{2}∠ACB = 22.5^{\circ}$,所以∠DAE = ∠BAC + ∠BAD + ∠CAE = 135°.
(2)∠DAE的度数不变.理由如下:因为∠BAC = 90°,所以∠ABC + ∠ACB = 90°.同(1)可得∠BAD = $\frac{1}{2}∠ABC$,∠CAE = $\frac{1}{2}∠ACB$,所以∠BAD + ∠CAE = $\frac{1}{2}(∠ABC + ∠ACB) = 45^{\circ}$,所以∠DAE = ∠BAD + ∠CAE + ∠BAC = 135°.
(3)∠DAE = 90° + $\frac{1}{2}∠BAC$

答案：
(1)因为∠BAC = 90°,AB = AC,所以∠ABC = ∠ACB = $\frac{1}{2}(180^{\circ} - ∠BAC) = 45^{\circ}$.因为DB = AB,EC = AC,所以∠BAD = ∠D,∠CAE = ∠E,所以∠ABC = ∠BAD + ∠D = 2∠BAD,∠ACB = ∠CAE + ∠E = 2∠CAE,所以∠BAD = $\frac{1}{2}∠ABC = 22.5^{\circ}$,∠CAE = $\frac{1}{2}∠ACB = 22.5^{\circ}$,所以∠DAE = ∠BAC + ∠BAD + ∠CAE = 135°.
(2)∠DAE的度数不变.理由如下:因为∠BAC = 90°,所以∠ABC + ∠ACB = 90°.同
(1)可得∠BAD = $\frac{1}{2}∠ABC$,∠CAE = $\frac{1}{2}∠ACB$,所以∠BAD + ∠CAE = $\frac{1}{2}(∠ABC + ∠ACB) = 45^{\circ}$,所以∠DAE = ∠BAD + ∠CAE + ∠BAC = 135°.
(3)∠DAE = 90° + $\frac{1}{2}∠BAC$