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Question
The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD . Prove that 2AB2 = 2AC2 + BC2.
Solution
Applying Pythagoras theorem for ΔACD, we obtain
AC2 = AD2 +DC2
AD2 = AC2 - DC2 ... (1)
Applying Pythagoras theorem in ΔABD, we obtain
AB2 = AD2 + DB2
AD2 = AB2 - DB2 ...(2)
From equation 1 and 2 we obtain
AC2 - DC2 = AB2 - DB2 ....(3)
it is given that 3DC = DB
`:.DC = (BC)/4 `
Putting these value in equation 3 we obtain
`(AC)^2 - ((BC)/4) ^2 = (AB)^2 - ((3BC)/4)^2`
`(AC)^2 - (BC)^2/16 = (AB)^2-`
16AC2 - BC2 = 16AB2 - 9BC2
16AB2 - 16AC2 8BC2
2AB2 = 2AC2 + BC2
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