Question

Hey guys
Sorry for disturbing you all at this time
but I need your help
Ques1:
ABC is a right triangle right angled at B and points D and E trisect BC. Prove that
8AE^2 = 3 AC ^2 + 5 AD^2
Refer the given attachment for figure .
Remember don't copy from any book or website .​

Answer 1

Answer:

Step-by-step explanation:

In Δ ABC,

BD = DE = EC = x

AC² = AB² + BC²  

(By Pythagoras theorem)

AC² = AB + (3x)²  (BC = CD + DE + EC = 3x)

AC² = AB² + 9x²

Now,

3AC² + 5AD² = 3(AB² + 9x²) + 5(AB² + x²)   [This is the RHS of what we have to prove]

8AB² + 32x²

8(AB² + 4x²)

= 8AE²  [thus LHS = RHS]

⇒ 8AE² = 3AC² + 5AD²

Hence proved.

Answer 2

Answer:

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#ASH