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Given right DABC, right angled at B. D and E are points of trisection of the side B
Let BD = DE = EC = k
Hence
we get BE = 2k and BC = 3k
In ΔABD, by Pythagoras theorem, we get
AD2 = AB2 + BD2 AD2 = AB2 + k2
Similarly, in ΔABE we get
AE2 = AB2 + BE2
Hence
AE2 = AB2 + (2k)2 = AB2+ 4k2
and AC2 = AB2 + BC2 = AB + (3k)2
AC2= AB2 + 9k2
Consider,
3AC2 + 5AD2
= 3(AB2 + 9k2) + 5(AB2 + 4k2)
= 8AB2 + 32k2= 8(AB2 + 4k2)
∴ 3AC2 + 5AD2 = 8AE2
Given right DABC, right angled at B. D and E are points of trisection of the side B
Let BD = DE = EC = k
Hence
we get BE = 2k and BC = 3k
In ΔABD, by Pythagoras theorem, we get
AD2 = AB2 + BD2 AD2 = AB2 + k2
Similarly, in ΔABE we get
AE2 = AB2 + BE2
Hence
AE2 = AB2 + (2k)2 = AB2+ 4k2
and AC2 = AB2 + BC2 = AB + (3k)2
AC2= AB2 + 9k2
Consider,
3AC2 + 5AD2
= 3(AB2 + 9k2) + 5(AB2 + 4k2)
= 8AB2 + 32k2= 8(AB2 + 4k2)
∴ 3AC2 + 5AD2 = 8AE2
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