Question

ABC is a right triangle right angled at B and points D and E trisect BC
Prove that 8 AE square = 3 AC square +5 AD square
​

Answer 1

Since D and E are the points of trisection of BC therefore BD=DE=CE

Let BD=DE=CE=x.

Then BE=2x and BC=3x

In rt △ABD,

AD

2

=AB

2

+BD

2

=AB

2

+x

2

(i)

In rt △ABE,

AE

2

=AB

2

+BE

2

=AB

2

+4x

2

(ii)

In rt △ABC,

AC

2

=AB

2

+BC

2

=AB

2

+9x

2

(iii)

Now, 8AE

2

−3AC

2

−5AD

2

=8(AB

2

+4x

2

)−3(AB

2

+9x

2

)−5(AB

2

+x

2

)

=0

Therefore, 8AE

2

=3AC

2

+5AD

2

Hi, hope it will be helpful