Question

Question #06bcd

Answer 1

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,

`AB^2 = AD^2 + BD^2 + 2 BD * DE` ...... `(i)`

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

Therefore,

`AC^2 = A^D2 + DC^2 - 2 DC * DE`

`AC^2 = AD^2 + BD^2 - 2 BD * DE` ........`(ii)`

(since, CD = BD)

Adding (i) and (ii)

`AB^2 + AC^2 = 2AD^2 + 2BD^2`

`AB^2 +AC^2 = 2(AD^2 + BD^2)`