the perpendicular from A on side BC of a triangle ABC intersect BC at D such that D B = 3 CD prove that 2AB²=2AC²+BC²
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Given that AD ⊥ BC and DB = 3CD
To prove : 2AB2 = 2AC2 + BC2
Proof :
BD + DC = BC
3CD + CD = BC
4CD = BC ⇒ CD = BC / 4
DB = 3CD = 3BC / 4.
In a right angle traingle ACD ,
AC2 = AD2 + CD2.
AC2 = AD2 + BC2 / 16 -------(1)
In a right angle traingle ABD ,
AB2 = AD2 + BD2.
AB2 = AD2 + 9BC2 / 16 -------(2).
Substracting (1) from (2) we obtain
AB2 - AC2 = 9BC2 / 16 - BC2 / 16
16(AB2 - AC2 ) = 8BC2
2(AB2 - AC2 ) = BC2
2AB2 = 2AC2 + BC2
To prove : 2AB2 = 2AC2 + BC2
Proof :
BD + DC = BC
3CD + CD = BC
4CD = BC ⇒ CD = BC / 4
DB = 3CD = 3BC / 4.
In a right angle traingle ACD ,
AC2 = AD2 + CD2.
AC2 = AD2 + BC2 / 16 -------(1)
In a right angle traingle ABD ,
AB2 = AD2 + BD2.
AB2 = AD2 + 9BC2 / 16 -------(2).
Substracting (1) from (2) we obtain
AB2 - AC2 = 9BC2 / 16 - BC2 / 16
16(AB2 - AC2 ) = 8BC2
2(AB2 - AC2 ) = BC2
2AB2 = 2AC2 + BC2
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