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

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Solution

Since $D$ and $E$ are the points of trisection of $B C$ therefore $B D = D E = C E$

Let $B D = D E = C E = x .$

Then $B E = 2 x$ and $B C = 3 x$

In rt $△ A B D,$

$A D^{2} = A B^{2} + B D^{2} = A B^{2} + x^{2}$ (i)

In rt $△ A B E,$

$A E^{2} = A B^{2} + B E^{2} = A B^{2} + 4 x^{2}$ (ii)

In rt $△ A B C,$

$A C^{2} = A B^{2} + B C^{2} = A B^{2} + 9 x^{2}$ (iii)

Now, $8 A E^{2} - 3 A C^{2} - 5 A D^{2}$

$= 8 (A B^{2} + 4 x^{2}) - 3 (A B^{2} + 9 x^{2}) - 5 (A B^{2} + x^{2})$

$= 0$

Therefore, $8 A E^{2} = 3 A C^{2} + 5 A D^{2}$

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