Question

AD is a median then prove that AB^2 + AC^2=2(AD^2 +BD^2)

Answer 1

Draw a perpendicular AL on BC

in ΔALD,

AD² = DL² + AL²

AL² = AD² - DL²

AGAIN IN ΔALB

AB² = AL² + BL²

AB² = AD² - DL² + BL²

AB² = AD² - DL² + (BD + DL)²

AB² = AD² - DL² + BD² + DL² + 2.BD.DL

AB² = AD² + BD² + 2.BD.DL................(I)

IN ΔALC

AC² = AL² + LC²

AC² = AD² - DL² + LC²

AC² = AD² - DL² + (DC - DL)²

AC² = AD² - DL² + DC² + DL² - 2.DC.DL

AC² = AD² + DC² - 2.DC.DL

AC² = AD² + BD² - 2.BD.DL .............................(II)

ADDING EQUATION I AND II

AC² + AB² = AD² +BD² - 2.BD.DL + AD² + BD² + 2.BD.DL

AC² + AB ² = 2(AD² + BD²)

proved

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Answer 2

in triangle ABC, AD is median

then , BD = BC

in triangles ABD and ADC

AB2 = AD2 + BD2 .........(1)

AC2 = AD2 + BC2 ............(2)

now add both the equations,

AB2 + AC2 = AD2 + AD2 + BD2+ BC2

AB2 + AC2 = 2AD2 + 2BD2 [ since BD = BC]

AB2 + AC2 = 2[AD2 + BD2]

hence proved...

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