题目

在△ABC中,∠C=90°,AC=BC=根号2+1,AD是∠CAB的角平分线,求CD的长

提问时间：2021-03-02

答案

过D作DE⊥AB
∵∠C=90°,AC=BC=√2+1
∴AB²＝AC²+BC²＝2（√2+1）²
∴AB＝√[2（√2+1）²]＝（√2+1）×√2
∵AD平分∠CAB,∠C=90°,DE⊥AB
∴CD＝DE
∴S△ACD＝AC×CD/2
S△ADB＝AB×DE/2＝AB×CD/2
S△ABC＝AC×BC/2
∵S△ABC＝S△ACD+ S△ADB
∴AC×CD/2+AB×CD/2＝AC×BC/2
AC×CD+ AB×CD＝AC×BC
（AC+AB）×CD＝AC×BC
[√2+1+（√2+1）×√2] ×CD＝（√2+1）²
（√2+1）²×CD＝（√2+1）²
CD＝1