$In a figure 𝐷, 𝐸 trisects the side 𝐵𝐶 of a triangle ∆𝐴𝐵𝐶 prove that 8𝐴𝐸² = 3𝐴𝐶² + 5𝐴𝐷²$

Given that,

 In a figure 𝐷, 𝐸 trisects the side 𝐵𝐶 of a triangle ∆𝐴𝐵𝐶

 Now to prove that, 8𝐴𝐸² = 3𝐴𝐶² + 5𝐴𝐷²

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, By using Pythagoras Theorem 

𝐴𝐷² = 𝐴𝐵² + 𝐵𝐷² 

𝐴𝐷² = 𝐴𝐵² + 𝑥² 

In Δ 𝐴𝐵𝐸,

 By using Pythagorus theorem 

𝐴𝐸² = 𝐴𝐵² + 𝐵𝐸² 

𝐴𝐸² = 𝐴𝐵² + (2𝑥)² 

𝐴𝐸² = 𝐴𝐵² + 4𝑥²

 In Δ 𝐴𝐵𝐶, 

By using Pythagorus theorem 

𝐴𝐶² = 𝐴𝐵² + 𝐵𝐶² 

𝐴𝐶² = 𝐴𝐵² + (3𝑥)² 

𝐴𝐶² = 𝐴𝐵² + 9𝑥²

Now, 𝑅𝐻𝑆 = 3𝐴𝐶² + 5𝐴𝐷²

 = 3(𝐴𝐵² + 9𝑥²) + 5(𝐴𝐵² + 𝑥²) 

= 8𝐴𝐵² + 32𝑥² 

=8(𝐴𝐵² + 4𝑥²) 

= 8𝐴𝐸² 

⇒ 8𝐴𝐸² = 3𝐴𝐶² + 5𝐴𝐷² 

Hence proved.