Math, asked by ananya2277, 3 months ago

In a triangle ABC, right angled at B,D and E trisect BC. Prove that 8AE^2 = 3AC^2 + 5AD^2.​

Answers

Answered by subhransusahoo94
6

Answer:

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

Answered by YOGESHmalik025
3

answer is attached above !

Attachments:
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