Question

If D and E trisect BC. The term 3AC2+5AD2=?​

Answer 1

Step-by-step explanation:

In ΔABC

D & E trisect BC.

Let AB=y

BD=x

DE=x & BC=3x, BE=2x

EC=x

Now, we get three right

angle triangle with help of Pythagoras

AD^2=AB^2+BD^2

AE^2=AB^2+BE^2

AC^2=AB^2+BC^2

Now, LHS=8AE^2

=8(AB^2+BE^2)

=8(y^2+(2x)^2)

LHS=8y^2+32x^2

RHS=3AC^2+5AD^2

RHS=3(AB^2+BC^2)+5(AB^2+BD^2)

RHS=3(y^2+9y^2)+5(x^2+y^2)

RHS=8y^2+32x^2

So, LHS=RHS

∴8AE^2=3AC^2+5AD^2

Hence Proved