Question

如图所示,AD是△ABC的中线。试判断AB²;+AC²;=2(AD²;+DC²)吗?并说明理由

急急急

Answer 1

证明：
做AE垂直于BC于E
则有 AB^2=BE^2+AE^2
AC^2=CE^2+AE^2
AE^2=AD^2-DE^2
因为 BD=CD
所以 BE=BD-DE=CD-DE
所以 CE=CD+DE
所以 AB^2+AC^2=(CD-DE)^2+(CD+DE)^2+2*(AD^2-DE^2)
=2*CD^2+2*DE^2+2*AD^2-2*DE^2
==2(AD^2+DC^2

Answer 2

做AE垂直于BC于E，则有
AB^2=BE^2+AE^2
AC^2=CE^2+AE^2
AE^2=AD^2-DE^2
BD=CD
BE=BD+DE=CD+DE
CE=CD-DE
AB^2+AC^2=(CD+DE)^2+(CD-DE)^2+2*(AD^2-DE^2)
=2*CD^2+2*DE^2+2*AD^2-2*DE^2
=2(AD^2+DC^2)