Question

in any triangle ABC prove that AB square +AC SQUARE =2(AD SQUARE +BD SQUARE), where D is mid point of BC

Answer 1

Sol:

In ΔABC, AB>AC, AD⊥BC,

According to Pythagoras theorem,
AB2 = AD2 + BD2
⇒ AB2 = AD2 + (BE + DE)2
⇒ AB2 = AD2 + BE2 + DE2 + 2 . BE . DE --------------------- (1)

AC2 = AD2 + CD2
AC2 = AD2 + (CE - DE)2
AC2 = AD2 + CE2+ DE2 - 2 . CE . DE --------------------- (2)

Adding equations (1) and (2),

AB2 + AC2 = AD2 + BE2 + DE2 + 2 . BE . DE + AD2 + CE2 + DE2 - 2 . CE . DE

But CE = BE,

AB2 + AC2 = 2AD2 + 2BE2 + 2DE2

⇒ AB2 + AC2 = 2(AD2 + DE2) + 2BE2

⇒ AB2 + AC2 = 2(AD2 + DE2) + 2BE2

⇒ AB2 + AC2 = 2AE2 + 2BE2 [AD2 + DE2 = AE2]

∴ AB2 + AC2 = 2AE2 + 2BE2

Hence proved.