《2026年练就优等生课后提分攻略九年级数学全一册人教版》

01 如图 5 - 78 - 1，$DE// BC$，且$EC:BD = 2:3$，$AD = 9$.
(1)求$AE$的长；
(2)若$AD:AB = 2:3$，$\triangle ADE$的面积是 4，求$\triangle ABC$的面积.

答案： 01 解：
(1)AE=6.
(2)
∵DE//BC,
∴△ADE∽△ABC.
∴$\frac{S_{\triangle ADE}}{S_{\triangle ABC}} = (\frac{AD}{AB})^2. $
∵AD:AB = 2:3, △ADE的面积是4,
∴$S_{\triangle ABC}=9.$

02 如图 5 - 78 - 2，在矩形$ABCD$中，$BE\perp AC$分别交$AC$，$AD$于点$F$，$E$，若$AE = 2$，$ED = 3$.
(1)求$\frac{EF}{BE}$的值；
(2)求$BE$的长.

答案： 02 解：
(1)
∵AE = 2, ED = 3,
∴AD = AE + ED = 5.
∵四边形ABCD是矩形,
∴BC = AD = 5, AD//BC.
∴△AEF∽△CBF.
∴$\frac{EF}{BF} = \frac{AE}{BC} = \frac{2}{5}. $
∵$\frac{EF}{BF + EF} = \frac{AE}{BC + AE} = \frac{2}{5 + 2} = \frac{2}{7}, $即$\frac{EF}{BE} = \frac{2}{7}. $
(2)设EF = 2x(x ＞ 0), 则BF = 5x, BE = 7x.
∵四边形ABCD是矩形,
∴∠BAE = 90°.
∵BE⊥AC,
∴∠AFE = 90°.
∴∠AFE = ∠BAE.
又
∵∠AEF = ∠BEA,
∴△AEF∽△BEA.
∴$\frac{AE}{BE} = \frac{EF}{AE}. $
∴$\frac{2}{7x} = \frac{2x}{2}. $
∴$x^2 = \frac{2}{7}. $
∵x ＞ 0,
∴$x = \frac{\sqrt{14}}{7}. $
∴$BE = 7x = \sqrt{14}.$

03 如图 5 - 78 - 3，已知$D$是$BC$的中点，$M$是$AD$的中点，求$AN:NC$的值.

答案： 03 解：如图, 过点D作AC的平行线交BN于点H.
∵DH//AC,
∴△BDH∽△BCN.
∴$\frac{DH}{CN} = \frac{BD}{BC}. $
∵D为BC的中点,
∴$\frac{DH}{CN} = \frac{BD}{BC} = \frac{1}{2}. $
∵DH//AN,
∴△DHM∽△ANM.
∴$\frac{DH}{AN} = \frac{DM}{AM}. $
∵M为AD的中点,
∴$\frac{DH}{AN} = \frac{DM}{AM} = 1. $
∴DH = AN.
∴$\frac{AN}{CN} = \frac{1}{2}, $即AN:NC = 1:2.