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