In Traingle ABC, if AD is the median, then show that AB^2 + AC^2 = 2(AD^2+BD^2)

Asked by jaychandran1967 | 27th Sep, 2016, 06:44: PM

Expert Answer:

Draw AE perpendicular to BC.

Since angle AED = 900,

Therefore, in triangle ADE

 angle ADE < 900 and angle ADB > 900

Thus, triangle ABD is obtuse angled triangle and triangle ADC is acute angled triangle.

ΔABD is obtuse angled at D and AE is perpendicular to BD produced,


AB2 = AD2 + BD2 + 2 BD x DE ……………(i)

Triangle ACD is acute angled at D and AE is perpendicular to BD produced,


AC2 = AD2 + DC2 - 2 DC x DE

 AC2 = AD2 + BD2 - 2 BD x DE  ………(ii)        (since CD = BD)

Adding (i) and (ii)

AB2 + AC2 = 2AD2 + 2BD2

AB2 + AC2 = 2(AD2 + BD2)

Answered by Rebecca Fernandes | 28th Sep, 2016, 08:55: AM

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