证明：
过A做AH⊥BC于H
根据勾股定理
AB²+AC²
=AH²+BH²+AH²+CH²
=2AH²+(BD-DH)²+(CD+DH)²
=2AH²+BD²+CD²-2BD*DH+2CD*DH
=2AH²+2DC²
如仍有疑惑,欢迎追问.祝：��֤��=2AD²+2DC²����不好意思，写漏了。AB²+AC²=AH²+BH²+AH²+CH²=2AH²+(BD-DH)²+(CD+DH)²=2AH²+BD²+CD²+2DH²-2BD*DH+2CD*DH=2AH²+2DH²+2DC²=2AD²+2DC²