Question

Prove the following .....

Answer 1

ABC is a triangle right angled at B, and D and E are points of trisection of BC.
Let BD = DE = EC = x
Then BE = 2x and BC = 3x
In Δ ABD,
AD² = AB² + BD² 
AD² = AB² + x²
In Δ ABE,
AE² = AB² + BE²
AE² = AB² + (2x)²
AE² = AB² + 4x²
In Δ ABC,
AC² = AB² + BC²  
AC² = AB + (3x)² 
AC² = AB² + 9x²
Now,
3AC² + 5AD² = 3(AB² + 9x²) + 5(AB² + x²) 
8AB² + 32x²
8(AB² + 4x²)
= 8AE²
⇒ 8AE² = 3AC² + 5AD²
Hence proved.