CBSE Class 10 Answered

We have, BD = DE = EC = BC/3

In ABD,

AD² = AB² + BD²

AD² = AB² + BC²/9       ..........BD = BC/3  ................(1)

Similarly other possible right angled triangles, we can write,

AE² = AB² + BE²

AE² = AB² + (2BC/3)²

AE² = AB² + 4BC²/9

and

AC² = AB² + BC²

AC² = AB² + BC²     ........(2)

Now,

AE² = AB² + 4BC²/9

8AE² = 8AB² + 32BC²/9

= 5AB² + 5xBC²/9 + 3AB² + 27BC²/9

= 5AB² + 5xBC²/9 + 3AB² + 3x9BC²/9

= 5(AB² + BC²/9) + 3(AB² + 9BC²/9)

= 5(AB² + BC²/9) + 3(AB² + BC²)

= 5AD² + 3AC²  ....................from (1) and (2)

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