Question

in a triangle ABC, B=right angle, points D and E trisect base BC then prove that 8 AE^2=3AC^2+5AD^2

Answer 1

Solution
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 right triangles ABD,ABE,ABC using Pythagoras theorem we have,
AD^2=AB²+x². (i)
AE^2=AB²+(2x)²
AE^2=AB²+4x². (ii)
AC^2=AB²+(3x)²
AC^2=AB²+9x². (iii)
L.H.S = 8AE²
putting the value obtained from (ii)
=8(AB²+4x²)
=8AB²+32x²
R.H.S=3AC²+5AD²
putting the value obtained from (i) and (iii)
=3(AB²+9x²)+5(AB²+x²)
=3AB²+27x²+5AB²+5x²
=8AB²+32x²
L.H.S.=R.H.S.

Answer 2

Step-by-step explanation:

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,

AE² =AB² +BE² =AB² +4x² (ii)

In rt △ABC,

AC² =AB² +BC² =AB² +9x² (iii)

Now, 8AE² −3AC² −5AD²

=8(AB² +4x²)−3(AB² +9x²)−5(AB² +x²)

=0

Therefore, 8AE² =3AC² +5AD²

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