《2025年名校题库八年级数学上册北师大版》

15.(高新区期末)如图,在$\triangle ABC$中,$\angle BAC= 90^{\circ}$,$AB= AC$.D是$\triangle ABC$所在平面内一点,且$\angle ADB= 90^{\circ}$.
(1)如图1,当点D在BC边上时,求证:$AD= CD$;
(2)如图2,当点D在$\triangle ABC$外部时,连接CD,若$AB= 5$,$AC= CD$,求线段BD的长;
(3)如图3,当点D在$\triangle ABC$内部时,连接CD,若$\angle ADC= \angle BDC$,$AD= 3$,求点D到BC的距离.

答案：
解:
(1)证明:$\because ∠BAC = 90^{\circ },AB = AC$，$\therefore ∠C = ∠B = 45^{\circ }$。$\because ∠ADB = 90^{\circ },\therefore ∠BAD = ∠DAC = 45^{\circ },\therefore ∠C = ∠DAC,\therefore AD = CD$。
(2)解:如图1，过点C作$CE⊥AD$于点E。

$\because AC = CD,\therefore AE = DE$。$\because ∠BAD + ∠EAC = 90^{\circ },∠EAC + ∠ACE = 90^{\circ },\therefore ∠BAD = ∠ACE$。又$\because AB = AC,∠ADB = ∠AEC,\therefore △ABD≌△CAE(AAS)$，$\therefore BD = AE,\therefore BD = AE = DE$。设$BD = x$，则$AD = 2x$。$\because ∠ADB = 90^{\circ },\therefore AD^{2} + BD^{2} = AB^{2}$，$\therefore (2x)^{2} + x^{2} = 5^{2}$，$\therefore x = \sqrt{5}$(负值已舍去)，$\therefore BD = \sqrt{5}$。
(3)解:如图2，过点C分别作$CN⊥AD$，交AD的延长线于点N，作$CM⊥BD$，交BD的延长线于点M，过点D作$DE⊥BC$于点E。

$\because ∠ADB = 90^{\circ },∠ADC = ∠BDC,\therefore ∠ADC = ∠BDC = \frac{1}{2}×(360^{\circ } - 90^{\circ }) = 135^{\circ },∠ADM = ∠BDN = 90^{\circ }$，$\therefore ∠CDM = ∠CDN = 45^{\circ },\therefore ∠MCD = ∠NCD = 45^{\circ },\therefore DM = CM = CN = DN$。与
(2)同理可证$△ABD≌△CAN$，$\therefore AD = CN = 3,\therefore BD = AN = 6,\therefore AB = \sqrt{AD^{2}+BD^{2}} = 3\sqrt{5}$，$\therefore BC = \sqrt{AB^{2}+AC^{2}} = 3\sqrt{10}$。$\because S_{△ADB} + S_{△ADC} + S_{△BDC} = S_{△ABC}$，$\therefore \frac{1}{2}×3×6 + \frac{1}{2}×3×3 + \frac{1}{2}×3\sqrt{10}\cdot DE = \frac{1}{2}×3\sqrt{5}×3\sqrt{5}$，$\therefore DE = \frac{3}{5}\sqrt{10}$，即点D到BC的距离为$\frac{3}{5}\sqrt{10}$。