Question

In triangle ABC, AD is median and AE is perpendicular to BC, prove that - AB^2 + AC^2 = 2AD^2 + 1/2BC^2

Answer 1

Answer:

its simple

Explanation:

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.

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 = 2AD2 + 2BD2

AB2 +AC2 = 2(AD2 + BD2

Answer 2

Explanation:

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.

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 = 2AD2 + 2BD2

AB2 +AC2 = 2(AD2 + BD2