《2025年新课程问题解决导学方案九年级数学上册华师大版》

21.(8 分)如图,在矩形 ABCD 中,E 为边 CD 上一点,且$AE \perp BD$.
(1)求证：$AD^{2} = DE \cdot DC$.
(2)F 为线段 AE 延长线上一点,且满足$EF = CF = \frac{1}{2}BD$,求证：$CE = AD$.

答案：
(1)$\because$四边形$ABCD$是矩形，$\therefore\angle BAD = 90^{\circ}$，$\angle ADE = 90^{\circ}$，$AB = DC$。$\therefore\angle ABD + \angle ADB = 90^{\circ}$。$\because AE\perp BD$，$\therefore\angle DAE + \angle ADB = 90^{\circ}$。$\therefore\angle ABD = \angle DAE$。$\because\angle BAD = \angle ADE = 90^{\circ}$，$\therefore\triangle ADE\backsim\triangle BAD$。$\therefore\frac{AD}{BA} = \frac{DE}{AD}$。$\therefore AD^{2} = DE\cdot BA$。$\because AB = DC$，$\therefore AD^{2} = DE\cdot DC$。
(2)如图，连结$AC$，交$BD$于点$O$。$\because$四边形$ABCD$是矩形，$\therefore\angle ADE = 90^{\circ}$。$\therefore\angle DAE + \angle AED = 90^{\circ}$。$\because AE\perp BD$，$\therefore\angle DAE + \angle ADB = 90^{\circ}$。$\therefore\angle ADB = \angle AED$。$\because\angle FEC = \angle AED$，$\therefore\angle ADO = \angle FEC$。$\because$四边形$ABCD$是矩形，$\therefore OA = OD = \frac{1}{2}BD$。$\therefore EF = CF = \frac{1}{2}BD$。$\therefore OA = OD = EF = CF$。$\therefore\angle ADO = \angle OAD$，$\angle FEC = \angle FCE$。$\because\angle ADO = \angle FEC$，$\therefore\angle ADO = \angle OAD = \angle FEC = \angle FCE$。在$\triangle ODA$和$\triangle FEC$中，$\begin{cases}\angle ODA = \angle FEC\\\angle OAD = \angle FCE\\OD = FE\end{cases}$$\therefore\triangle ODA\cong\triangle FEC(A.A.S.)$。$\therefore CE = AD$。

22.(8 分)如图,在$\triangle ABC$中,$AB = AC = 1$,$BC = \frac{\sqrt{5} - 1}{2}$,在边 AC 上截取$AD = BC$,连结 BD.
(1)通过计算判断$AD^{2}与AC \cdot CD$的大小关系.
(2)求$∠ABD$的度数.

答案：
(1)$\because AD = BC$，$BC = \frac{\sqrt{5}-1}{2}$，$\therefore AD = \frac{\sqrt{5}-1}{2}$，$DC = 1 - \frac{\sqrt{5}-1}{2} = \frac{3 - \sqrt{5}}{2}$。又$AD^{2} = \left(\frac{\sqrt{5}-1}{2}\right)^{2} = \frac{3 - \sqrt{5}}{2}$，$AC\cdot CD = 1×\frac{3 - \sqrt{5}}{2} = \frac{3 - \sqrt{5}}{2}$，$\therefore AD^{2} = AC\cdot CD$。
(2)$\because AD = BC$，$AD^{2} = AC\cdot CD$，$\therefore BC^{2} = AC\cdot CD$，即$\frac{BC}{CD} = \frac{AC}{BC}$。又$\because\angle C = \angle C$，$\therefore\triangle ACB\backsim\triangle BCD$。$\therefore\frac{AB}{BD} = \frac{AC}{BC}$，$\angle A = \angle DBC$。$\because AB = AC$，$BC = AD$，$\therefore\frac{AB}{AC} = \frac{BD}{CB} = 1$。$\therefore DB = CB = AD$。$\therefore\angle A = \angle ABD$，$\angle C = \angle BDC$。设$\angle A = x$，则$\angle ABD = x$，$\angle DBC = x$，$\angle C = 2x$。$\because\angle A + \angle ABC + \angle C = 180^{\circ}$，$\therefore x + 2x + 2x = 180^{\circ}$。解得$x = 36^{\circ}$。$\therefore\angle ABD = 36^{\circ}$。