Question

Answer these questions

Answer 1

1.
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