Question
In the adjacent figure, $A B C$ is a right angled triangle with right angle at $B$ and points $D, E$ trisect $B C$. Prove that $8 A E^2$ $=3 A C^2+5 A D^2$
✓
Answer
Since Points D, E trisect BC.
BD = DE = CE
Let BD = DE = CE = x
BE = 2x and BC = 3x
In the right $\triangle A B D$,
$A D^2=A B^2+B D^2$
$A D^2=A B^2+x^2 \ldots(1)$
In the right $\triangle A B E$,
$A E^2=A B^2+2 B E^2$
$A E^2=A B^2+4 x^2 \ldots(2) \ldots(B E=2 x)$
In the right $\triangle A B C$
$A C^2=A B^2+B C^2$
$A C^2=A B^2+9 x^2 \ldots(3) \ldots(B C=3 x)$
$\text { R.H.S. }=3 A C^2+5 A D^2$
$=3\left[A B^2+9 x^2\right]+5\left[A B^2+x^2\right] \ldots[\text { From (1) and (3)] }$
$=3 A B^2+27 x^2+5 A B^2+5 x^2$
$=8 A B^2+32 x^2$
$=8\left(A B^2+4 x^2\right)$
$=8 A E^2 \ldots[\text { From (2)] }$
$=\text { R.H.S. }$
$\therefore 8 A E^2=3 A C^2+5 A D^2$
BD = DE = CE
Let BD = DE = CE = x
BE = 2x and BC = 3x
In the right $\triangle A B D$,
$A D^2=A B^2+B D^2$
$A D^2=A B^2+x^2 \ldots(1)$
In the right $\triangle A B E$,
$A E^2=A B^2+2 B E^2$
$A E^2=A B^2+4 x^2 \ldots(2) \ldots(B E=2 x)$
In the right $\triangle A B C$
$A C^2=A B^2+B C^2$
$A C^2=A B^2+9 x^2 \ldots(3) \ldots(B C=3 x)$
$\text { R.H.S. }=3 A C^2+5 A D^2$
$=3\left[A B^2+9 x^2\right]+5\left[A B^2+x^2\right] \ldots[\text { From (1) and (3)] }$
$=3 A B^2+27 x^2+5 A B^2+5 x^2$
$=8 A B^2+32 x^2$
$=8\left(A B^2+4 x^2\right)$
$=8 A E^2 \ldots[\text { From (2)] }$
$=\text { R.H.S. }$
$\therefore 8 A E^2=3 A C^2+5 A D^2$
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