Question

In the given figure, D and E trisect BC. Prove that 8AE2=3AC2+5AD2

Answer 1

Given:

∠B = 90°

D and E trisect BC

To Prove:

8AE² = 3AC² + 5AD²

Proof:

D and E trisect BC,

∴ BD = DE = EC = x

In Δ ABC

By using Pythagoras Theorem,

AC² = AB² + BC²

AC² = AB² + [BD + DE + EC]²

AC² = AB² + [x + x + x]²  (given)

AC² = AB² + [3x]²

AC² = AB² + 9x²

Multiply the above equation by 3.

3AC² = 3AB² + 27x² → Eq1

In Δ ABE

By using Pythagoras Theorem,

AE² = AB² + BE²

AE² = AB² + [BD + DE]²

AE² = AB² + [x + x]²  (given)

AE² = AB² + [2x]²

AE² = AB² + 4x²

Multiply the above equation by 8.

8AE² = 8AB² + 32x² → Eq2

In Δ ABD

By using Pythagoras Theorem,

AD² = AB² + BD²

AD² = AB² + x² (given)

Multiply the above equation by 5.

5AD² = 5AB² + 5x² → Eq3

Adding Equation 1 and Equation 3

3AC² + 5AD² = 3AB² + 5AB² + 27x² + 5x²

3AC² + 5AD² = 8AB² + 32x²

Apply Equation 2

∴ 3AC² + 5AD² = 8AE²

8AE² = 3AC² + 5AD²

Hence Proved.

Answer 2

Answer:

HOPE IT HELPS U...

Step-by-step explanation:

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.

   NOW MARK AS BRAINLIEST ANSWER