Question

In a right angled triangle ABC if D and E trisect BC then prove that $8A{E^2} = 3A{C^2} + 5A{D^2}$.

Answer 1

Hint- Since, it is a right angled triangle, make use of the Pythagoras theorem and try to solve this question.

Complete step-by-step answer:

Using the data given first let us try to draw the figure

Now, in the data it is given that D and E are the points of trisection,
So, if they are the points of trisection, we can write
BD=DE=CE
Let us consider BD=DE=CE=x
Then BE=2x, and BC=3x
Now let us consider right angled triangle ABD,
Let us apply the Pythagoras theorem to this triangle
So, we get
$A{D^2} = A{B^2} + B{D^2} = A{B^2} + {x^2}$ -----(i)
Similarly in right angled triangle ABE, on applying Pythagoras theorem we can write
$A{E^2} = A{B^2} + B{E^2} = A{B^2} + 4{x^2}$ ----- (ii)
Similarly in right angled triangle ABC,
Let us apply Pythagoras theorem
So, we can write
$A{C^2} = A{B^2} + B{C^2} = A{B^2} + 9{x^2}$ --------(iii)
Now, in the question we have been asked to prove that $8A{E^2} = 3A{C^2} + 5A{D^2}$
So, making use of the values we got in eq(i), eq(ii), eq(iii)
We can write
$8A{E^2}$
$
   = 8(A{B^2} + 4{x^2}) \\
   = 8A{B^2} + 32{x^2} \\
$
$
  3A{C^2} = 3(A{B^2} + 9{x^2}) \\
  3A{C^2} = 3A{B^2} + 27{x^2} \\
$
$
  5A{D^2} = 5(A{B^2} + {x^2}) \\
  5A{D^2} = 5A{B^2} + 5{x^2} \\
$
Now adding $3A{C^2} + 5A{D^2}$ ,
We get $8A{B^2} + 32{x^2}$ =$8A{E^2}$
So, from this we have proved that $8A{E^2} = 3A{C^2} + 5A{D^2}$
Hence, the result is proved.

Note: Here, in this question it is given to us that these are the points of trisection, in case we were given the points of bisection, then we would have only two points on the line BC. So, in accordance with the type of intersection given, solve the problem.