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Theorem of Geometric Mean

In the given ...

Question

In the given figure, $A B C$ is a triangle right angled at $B . D$ and $E$ are points on $B C$ trisect it.Prove that $8 {A E}^{2} = 3 {A C}^{2} + 5 {A D}^{2}$

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Solution

Given right angle $Δ A B C$ , right-angled at $B$ .

$D$ and $E$ are points of trisection of the side $B C$

Let $B D = D E = E C = k$

Hence we get $B E = 2 k$ and $B C = 3 k$

In $△ A B D$ , by Pythagoras theorem, we get

$A D^{2} = A B^{2} + B D^{2}$

$A D^{2} = A B^{2} + k^{2}$

In $△ A B E$ we get

$A E^{2} = A B^{2} + B E^{2}$

Hence $A E^{2} = A B^{2} + {(2 k)}^{2}$

$= A B^{2} + 4 k^{2}$ and

$A C^{2} = A B^{2} + B C^{2}$

$= A B + {(3 k)}^{2}$

$A C^{2} = A B^{2} + 9 k^{2}$

Consider, $3 A C^{2} + 5 A D^{2} = 3 (A B^{2} + 9 k^{2}) + 5 (A B^{2} + 4 k^{2})$

$= 8 A B^{2} + 32 k^{2}$

$= 8 (A B^{2} + 4 k^{2})$

$∴ 3 A C^{2} + 5 A D^{2} = 8 A E^{2}$

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