Question

If D and E trisect BC prove that 8AE^2=3AC^2+5AD^2

Answer 1

Answer:

Diagram of this question attached in attachment file.

Given right DABC, right angled at B.

D and E are points of trisection of the side BC

Let BD = DE = EC = k

Hence we get BE = 2k and BC = 3k

In ΔABD, by Pythagoras theorem, we get

AD2 = AB2 + BD2

AD2 = AB2 + k2

Similarly, in ΔABE we get

AE2 = AB2 + BE2

Hence AE2 = AB2 + (2k)2

= AB2 + 4k2 and

AC2 = AB2 + BC2

= AB + (3k)2

AC2= AB2 + 9k2

Consider, 3AC2 + 5AD2 = 3(AB2 + 9k2) + 5(AB2 + 4k2)

= 8AB2 + 32k2

= 8(AB2 + 4k2)

∴ 3AC2 + 5AD2 = 8AE2

Answer 2

since D and E trisect BC

BD=DE=EC

Let BD=DE=DC

Then

BE=2x and BC=3x

In triangle ABD

AD^2=AB^2+BD^2

AD^2=AB^2+x^2 ---------(1)

In triangle ABE

AE^2=AB^2+4x^2--------(2)

In triangle ABC

AC^2=AB^2+9x^2--------(3)

NOW,

8AE^2-3AC^2-5AD^2

8(AB^2+4x^2)-3(AB^2+9x^2)-5(AB^2+x^2)

8AB^2+32x^2-3AB^2-27x^2-5AB^2-5x^2=0

=8AE^2=3AC^2+5AD^2

HENCE PROVED