Question

In an acute angled triangle ABC, AD is the median then prove that AD2 = AB^2/2 + AC^2/2 - BC^2/4

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:Draw AE perpendicular to BC.

Since angle AED = 900,

Therefore, in triangle ADE

angle ADE 900

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

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

Therefore,

AB2 = AD2 + BD2 + 2 BD x DE ?????(i)

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

Therefore,

AC2 = AD2 + DC2 - 2 DC x DE

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

Adding (i) and (ii)

AB2 + AC2 = 2(AD2 + 2BD2)

AB2 +AC2 = 2(AD2 + BD2)