Question

In the given figure D and E trisect BC.If BD=DE=EC.Prove that 8AE×AE=3AC×AC+5AD×AD

Answer 1

Hence proved 8AE×AE=3AC×AC+5AD×AD

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 8AE×AE=3AC×AC+5AD×AD

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P is the midpoint of side BC of parallelogram ABCD such that angle BAP= angle DAP. Prove that AD= 2 times CD. ?

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