《2025年名校题库九年级数学全一册北师大版》

8. (锦江区一诊)如图1,在$□ ABCD$中,$E$,$F分别为DC$,$AB$的中点,连接$AE$,$CF$,且$\angle FCB= \angle FBC$.
(1)求证：四边形$AFCE$是菱形.
(2)如图2,连接$BD交AE于点G$,交$CF于点H$,且$CF\perp BD$,连接$CG$,$CA$.
①求证：$\angle CGB= \angle ACF$；
②若$CG= \sqrt{6}$,求$AB$的长.

答案：
(1)证明：$\because$四边形$ABCD$是平行四边形，
$\therefore AB// CD$，$AB = CD$。
$\because E$，$F$分别为$DC$，$AB$的中点，$\therefore CE = \frac{1}{2}CD$，$AF = BF = \frac{1}{2}AB$，
$\therefore CE = AF$，
$\therefore$四边形$AFCE$是平行四边形。
$\because \angle FCB = \angle FBC$，$\therefore CF = BF$，$\therefore CF = AF$，
$\therefore$四边形$AFCE$是菱形。
(2)①证明：$\because CF = AF = BF = \frac{1}{2}AB$，
$\therefore \angle ACB = 90^{\circ}$。
$\because$四边形$AFCE$是菱形，$\therefore AE// CF$，
$\therefore BH = HG$。
$\because CF\perp BD$，$\therefore CG = BC$，$\therefore \angle BCH = \angle GCH$。
$\because \angle BCH + \angle ACF = \angle CGH + \angle GCH = 90^{\circ}$，$\therefore \angle CGB = \angle ACF$。
②解：$\because CD// AB$，$\therefore \triangle CDH\backsim\triangle FBH$，
$\therefore \frac{CH}{FH} = \frac{CD}{BF}$。
$\because CD = AB = 2BF$，$\therefore \frac{CH}{FH} = 2$。
设$FH = x$，$CH = 2x$，$\therefore AF = BF = CF = 3x$，
$\therefore AB = 6x$。
$\because AE// CF$，$\therefore \angle AGB = \angle FHB = 90^{\circ}$，$AG = 2FH = 2x$，
$\therefore BG = \sqrt{AB^{2} - AG^{2}} = 4\sqrt{2}x$，
$\therefore HG = \frac{1}{2}BG = 2\sqrt{2}x$。
$\because CG^{2} = CH^{2} + HG^{2}$，$\therefore 6 = 4x^{2} + (2\sqrt{2}x)^{2}$，解得$x = \frac{\sqrt{2}}{2}$（舍去负值），$\therefore AB = 6x = 3\sqrt{2}$。

1. (七中育才)如图,在$Rt△ABC$中,$∠BAC= 90^{\circ },AD⊥BC$于点 D.若$\frac {AD}{CD}= \frac {3}{2}$,则$\frac {AB}{AC}= $(A)

A. $\frac {3}{2}$
B. $\frac {2}{3}$
C. $\frac {\sqrt {6}}{2}$
D. $\frac {\sqrt {6}}{3}$

答案： A

2. (锦江区一诊)如图,在$△ABC$中,点 D 在线段 BC 上,且$∠BAD= ∠C,BD= 2,CD= 6$,则 AB 的长是 (C)

A. 12
B. 8
C. 4
D. 3

答案： C