推荐回答（3个）
1、画图可以知道有直角三角形ABD且角B=60度
角BAD=30度
所以AB=2BD=3分之2根号3
同时角C=45度且有直角三角形ADC
所以AD=DC=1，由勾股定理得AC^2=AD^2+DC^2
所以AC=根号2
2、（1）因为直角三角形ABC所以AB^2+AC^2=BC^2
同时有直角三角形ABD、ADC
所以BD^2=AB^2-AD^2
CD^2=AC^2-AD^2
所以BC^2=AB^2+AC^2=AB^2-AD^2+AC^2-AD^2+2AD^2
所以AB^2+AC^2=AB^2+AC^2
所以得证
（2）只要证明三角形ABD∽三角形ADC
即有AD/CD=BD/AD
所以AD^2=BD*DC
1. AD/BD=tan∠B，AD=BDtan∠B=1
AB=BD/cos∠b=2 根号3/3
AC=CD/cos∠C=根号2
2.BC²=AB²+AC²=BD²+DC²+2AD²
(BD+DC)²=BC²=BD²+DC²+2AD²
2BD*DC=2AD²
BD*DC=AD²
1、AD=DC=1
AB=2BD=2根号3/3，
AC=根号2DC=根号2
2、由勾股定理得，BC²=AB²+AC²,又因为AD垂直BC于D，所以AB²=BD²+AD²,AC=AD²+DC²,将这两个式子带入上式得BC²=BD²+DC²+2AD²。
（2）因为三角形ABD和三角形CAD是相似三角形，所以AD/BD=DC/AD,所以AD²=BD乘DC
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