《2025年全练单元卷九年级数学下册人教版》

26. (10分)如图，$\triangle ABC$中，$\angle ABC=90^{\circ}$，$F$是$AC$的中点，过$AC$上一点$D$作$DE// AB$，交$BF$的延长线于点$E$，$AG\perp BE$，垂足是$G$，连接$BD$，$AE$.
(1)求证：$\triangle ABC\sim\triangle BGA$；
(2)若$AF = 5$，$AB = 8$，求$FG$的长；
(3)当$AB = BC$，$\angle DBC = 30^{\circ}$时，求$\frac{DE}{BD}$的值.

答案：

(1)证明：$\because \angle ABC = 90^{\circ}$，$F$是$AC$的中点，
$\therefore BF=\frac{1}{2}AC = AF$，
$\therefore \angle FAB = \angle FBA$，
$\because AG\perp BE$，
$\therefore \angle AGB = 90^{\circ}$，
$\therefore \angle ABC = \angle AGB$，
$\therefore \triangle ABC\sim\triangle BGA$；
(2)解：$\because AF = 5$，
$\therefore AC = 2AF = 10$，$BF = 5$，
$\because \triangle ABC\sim\triangle BGA$，
$\therefore \frac{AB}{BG}=\frac{AC}{BA}$，$\therefore BG=\frac{AB^{2}}{AC}=\frac{8^{2}}{10}=\frac{32}{5}$，
$\therefore FG = BG - BF=\frac{32}{5}-5=\frac{7}{5}$；
(3)解：延长$ED$交$BC$于$H$，
则$DH\perp BC$，
$\therefore \angle DHC = 90^{\circ}$，
$\because AB = BC$，$F$为$AC$的中点，
$\therefore \angle C = 45^{\circ}$，$\angle CBF = 45^{\circ}$，
$\therefore \triangle DHC$，$\triangle BEH$是等腰直角三角形，
$\therefore DH = HC$，$EH = BH$，
设$DH = HC = a$，
$\because \angle DBC = 30^{\circ}$，
$\therefore BD = 2a$，$\therefore BH = \sqrt{3}a$，
$\therefore EH = \sqrt{3}a$，
$\therefore DE = (\sqrt{3}-1)a$，
$\therefore \frac{DE}{BD}=\frac{(\sqrt{3}-1)a}{2a}=\frac{\sqrt{3}-1}{2}$.