《2025年新课程问题解决导学方案九年级数学上册华师大版》

23.(12 分)如图,在四边形 ABCD 中,$AB = AC = AD$,AC 平分$∠BAD$,P 是 AC 延长线上一点,且$PD \perp AD$.
(1)求证：$∠BDC = ∠PDC$.
(2)若 AC 与 BD 相交于点 E,$AB = 1$,$CE:CP = 2:3$,求 AE 的长.

答案：
(1)$\because AB = AD$，$AC$平分$\angle BAD$，$\therefore AC\perp BD$。$\therefore\angle ACD + \angle BDC = 90^{\circ}$。$\because AC = AD$，$\therefore\angle ACD = \angle ADC$。$\therefore\angle ADC + \angle BDC = 90^{\circ}$。$\because PD\perp AD$，$\therefore\angle ADC + \angle PDC = 90^{\circ}$。$\therefore\angle BDC = \angle PDC$。
(2)如图，过点$C$作$CM\perp PD$于点$M$。$\because\angle BDC = \angle PDC$，$\therefore CE = CM$。$\because\angle CMP = \angle ADP = 90^{\circ}$，$\angle P = \angle P$，$\therefore\triangle CPM\backsim\triangle APD$。$\therefore\frac{CM}{AD} = \frac{PC}{PA}$。设$CM = CE = x$，$\because CE:CP = 2:3$，$\therefore PC = \frac{3}{2}x$。$\because AB = AD = AC = 1$，$\therefore\frac{x}{1} = \frac{\frac{3}{2}x}{\frac{3}{2}x + 1}$。解得$x = \frac{1}{3}$。$\therefore AE = 1 - \frac{1}{3} = \frac{2}{3}$。