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9. 如图,$Rt\triangle AB'C'是由Rt\triangle ABC绕点A$按顺时针方向旋转得到的,连接$CC'交AB于点E$,$CC'的延长线交BB'于点F$.
(1) 证明：$\triangle ACE\backsim \triangle FBE$；
(2) 设$\angle ABC= \alpha,\angle CAC'= \beta$,试探索$\alpha,\beta$满足什么关系时,$\triangle ACE与\triangle FBE$是全等三角形,并说明理由.
(1)证明：$\because Rt\triangle AB'C'$ 是由 $Rt\triangle ABC$ 绕点 $A$ 顺时针旋转得到的，$\therefore AC = AC'$，$AB = AB'$，$\angle CAB = \angle C'AB'$。$\therefore \angle CAC' = \angle BAB'$。$\therefore \angle ACC' = \angle ABB'$。又 $\angle AEC = \angle FEB$，$\therefore \triangle ACE \backsim \triangle FBE$。(2)解：当
(1) 证明：$\triangle ACE\backsim \triangle FBE$；
(2) 设$\angle ABC= \alpha,\angle CAC'= \beta$,试探索$\alpha,\beta$满足什么关系时,$\triangle ACE与\triangle FBE$是全等三角形,并说明理由.
(1)证明：$\because Rt\triangle AB'C'$ 是由 $Rt\triangle ABC$ 绕点 $A$ 顺时针旋转得到的，$\therefore AC = AC'$，$AB = AB'$，$\angle CAB = \angle C'AB'$。$\therefore \angle CAC' = \angle BAB'$。$\therefore \angle ACC' = \angle ABB'$。又 $\angle AEC = \angle FEB$，$\therefore \triangle ACE \backsim \triangle FBE$。(2)解：当
$\beta = 2\alpha$
时，$\triangle ACE \cong \triangle FBE$。在 $\triangle ACC'$ 中，$\because AC = AC'$，$\therefore \angle ACC' = \frac{180^{\circ} - \angle CAC'}{2} = \frac{180^{\circ} - \beta}{2} = 90^{\circ} - \alpha$。在 $Rt\triangle ABC$ 中，$\angle ACC' + \angle BCE = 90^{\circ}$，即 $90^{\circ} - \alpha + \angle BCE = 90^{\circ}$，$\therefore \angle BCE = \alpha$。$\because \angle ABC = \alpha$，$\therefore \angle ABC = \angle BCE$。$\therefore CE = BE$。由(1)知：$\triangle ACE \backsim \triangle FBE$，$\therefore \triangle ACE \cong \triangle FBE$。答案
(1)证明：$\because Rt\triangle AB'C'$ 是由 $Rt\triangle ABC$ 绕点 $A$ 顺时针旋转得到的，$\therefore AC = AC'$，$AB = AB'$，$\angle CAB = \angle C'AB'$。$\therefore \angle CAC' = \angle BAB'$。$\therefore \angle ACC' = \angle ABB'$。又 $\angle AEC = \angle FEB$，$\therefore \triangle ACE \backsim \triangle FBE$。(2)解：当 $\beta = 2\alpha$ 时，$\triangle ACE \cong \triangle FBE$。在 $\triangle ACC'$ 中，$\because AC = AC'$，$\therefore \angle ACC' = \frac{180^{\circ} - \angle CAC'}{2} = \frac{180^{\circ} - \beta}{2} = 90^{\circ} - \alpha$。在 $Rt\triangle ABC$ 中，$\angle ACC' + \angle BCE = 90^{\circ}$，即 $90^{\circ} - \alpha + \angle BCE = 90^{\circ}$，$\therefore \angle BCE = \alpha$。$\because \angle ABC = \alpha$，$\therefore \angle ABC = \angle BCE$。$\therefore CE = BE$。由(1)知：$\triangle ACE \backsim \triangle FBE$，$\therefore \triangle ACE \cong \triangle FBE$。
10. 某游戏软件公司开发出一种游戏软件,前期投入的开发广告宣传费用共$50000$元,且每售出一套软件,软件公司还需支付安装调试费用$200$元.
(1) 试写出总费$y$(元)与销售套数$x$(套)之间的函数关系式.
(2) 如果每套定价$700$元,软件公司至少要售出多少套软件才能确保不亏本.
(1) 试写出总费$y$(元)与销售套数$x$(套)之间的函数关系式.
(2) 如果每套定价$700$元,软件公司至少要售出多少套软件才能确保不亏本.
答案
(1)$y = 50000 + 200x$。(2)设软件公司至少要售出 $x$ 套软件才能保证不亏本，则有 $700x \geq 50000 + 200x$，解得 $x \geq 100$。答：软件公司至少要售出 100 套软件才能确保不亏本。
