Q. 163.8( 8 Votes )

In an acute-angle

Answer :

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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

 

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