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Question
The perpendicular from A on side BC of a Δ ABC meets 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 equation (2), we obtain
AC2 - DC2 = AB2 - DB2 .....(3)
It is given that 3DC = DB
∴ `"DC" = "BC"/4` and `"DB"=(3"BC")/ 4`
Putting these values in equation (3), we obtain
`"AC"^2 - ("BC"/4)^2 = "AB"^2 - ((3"BC")/4)^2`
`"AC"^2 - "BC"^2/16 = "AB"^2 - (9"BC"^2)/4`
16AC2 - BC2 = 16AB2 - 9BC2
16AB2 - 16AC2 = 8BC2
2AB2 = 2AC2 + BC2
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