In the figure, $D$ and $E$ trisect $B C$ , Prove that $8 {A E}^{2} = 3 {A C}^{2} + 5 {A D}^{2}$
[4 Marks]

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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 $B C = 3 x$ .
(1 Mark)

In right triangles $Δ A B D, Δ A B E$ and $Δ A B C$ , we have

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

$\Rightarrow {A D}^{2} = {A B}^{2} + x^{2}$

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

$\Rightarrow {A E}^{2} = {A B}^{2} + 4 x^{2}$

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

$\Rightarrow {A C}^{2} = {A B}^{2} + 9 x^{2}$
(1.5 Marks)

Now consider,

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

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

$\Rightarrow 8 {A E}^{2} - 3 {A C}^{2} - 5 {A D}^{2} = 0$

$\Rightarrow 8 {A E}^{2} = 3 {A C}^{2} + 5 {A D}^{2}$
(1.5 Marks)

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Similar questions

In the figure, D and E trisect BC, Prove that $8 {A E}^{2} = 3 {A C}^{2} + 5 {A D}^{2}$ [4 MARKS]

In the figure $∠ B = 90^{\circ}$ , D and E trisect BC, Prove that $8 {A E}^{2} = 3 {A C}^{2} + 5 {A D}^{2}$ [4 MARKS]

D

E

B C

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

∠ B = 90^{\circ}

B C

D

E

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

8 {A E}^{2} = 3 {A C}^{2} + 5 {A D}^{2} .

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