《2025年思维新观察八年级数学下册人教版天津专版》

1.如图，已知$\angle ACB = \angle CDB = 90^{\circ}$，$CD = BD$，$AC = 2$，$BC = 4$，求$AD$的长.
　　

答案： 作$AE\perp CD$于点$E$，$AE = \sqrt{2}$，$CD = 2\sqrt{2}$，$CE = \sqrt{2}$，$AD = \sqrt{2 + 18}=2\sqrt{5}$。

2.如图，已知$\angle ACB = \angle ADC = 90^{\circ}$，$AD = CD$，$AC = 2$，$BC = 4$，求$BD$的长.
　　

答案： 过$B$点作$BE\perp CD$于$E$点，$CD = \sqrt{2}$，$CE = BE = 2\sqrt{2}$，$BD = \sqrt{18 + 8}=\sqrt{26}$。

3.如图，已知$\angle ACB = \angle DAB = 90^{\circ}$，$AD = AB$，$AC = 2$，$BC = 4$，求$CD$的长.
　　　　　　　

答案： 过$D$点作$DF\perp AC$于点$F$，易证$\triangle ACB\cong\triangle DFA$，$\therefore CD=\sqrt{6^{2}+2^{2}} = 2\sqrt{10}$。

4.如图，已知$\angle ACB = \angle ABD = 90^{\circ}$，$AB = BD$，$AC = 2$，$BC = 4$，求$CD$的长.
　　

答案： 过$D$点作$DF\perp CB$于点$F$，易证$\triangle CAB\cong\triangle FBD$，$\therefore CD=\sqrt{36 + 16}=2\sqrt{13}$。

5.如图，已知$\angle ACB = \angle ADB = 90^{\circ}$，$AD = BD$，$AC = 2$，$BC = 4$，求$CD$的长.
　　

答案： 过$D$点$DE\perp CD$，交$CB$的延长线于点$E$，易证$\triangle ACD\cong\triangle BED$，$CD = 3\sqrt{2}$。

6.如图，已知$\angle ACB = \angle ADB = 90^{\circ}$，$AD = BD$，$AC = 2$，$BC = 4$，求$CD$的长.
　　

答案： 作$DE\perp CD$交$BC$于$E$点，易证$\triangle ACD\cong\triangle BED$，$\therefore BE = 2$，$\therefore CD=\sqrt{2}$。