In an acute-angle

_{We have}

In ABC, AD is median

AE⊥BC

In AEB

AB²=AE²+BE²

AB²=AD²-DE²+(BD-DE)²

AB²=AD²-DE²+BD²-2xBDxDE+DE²

AB²=AD²+BD²-2xBDxDE

AB²=AD²+BC²/4-BCxDE …………. (I) [GIVEN BC=2BD]

In AEC

AC²=AE²+EC²

AC²=AD²-DE²+ (DE+CD)²

AC²=AD²-DE²+2CDxDE

AC²=AD²+BC²/4+BCxDE ……….(II) [BC=2CD]

By adding equ. (i) and (ii) we get

AB²+AC²=2AD²+BC²/2

2AB²+2AC²=4AD²+BC² [MULTIPLY BY 2]

4AD²=2AB²+2AC²-BC²

AD²=2AB²+2AC²-BC²