Question #06bcd

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Answer 1 · 2015-09-13T08:19:23

its simple

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)

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

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