(Q20) In the given figure, ABC is a triangle right angled at B. D and E are points on BC trisect it. Prove that 8AE2 = 3AC2 + 5AD2.

Solution :

ABCDE

In ∆ABC, ∠B = 90°

⇒ AD is hypotenuse.

AC2 = AB2 + BC2

3AC2 = 3AB2 + 3BC2.....(1)

In ∆ABC, ∠B = 90°

⇒ AD is hypotenuse.

∴ AD2 = AB2 + BD2 = AB2 + (
BC
3
 )2
⇒ 2 = 2 +
2
⇒ 52 = 52 +
52
 .....(2)

(1) + (2)

AC2 + 52 = 32 + 3BC2 + 52 +
BC2 = 82 +
BC2.....(3)

Now in ∆ABE, ∠B = 90°

⇒ AE is hypotenuse.

⇒ AE2 = AB2 + BE2 = AB2 + (
2
3
 BC)2
= AB2 +
4
9
 BC2
⇒ AE2 = 8AB2 +
32
9
 BC2.....(4)

∴ RHS of (3) and (4) are .

∴ LHS of (3) and (4) are .

∴ AE2 = AC2 + AD2.

Hence Proved.