In the figure shown above, $△ A B C$ is a right-angled triangle which is right-angled at $B$ and $D$ and $E$ are two points on BC such that they trisect $B C$ as shown in the figure. Prove that $8 A E^{2} = 3 A C^{2} + 5 A D^{2}$ .

Open in App

Solution

REF.Image
Let $A B = y$ and $B C = 3 x$
$∴ B D = D E = C E = x$
$∴ B E = 2 x$

To Prove : $8 A E^{2} = 3 A C^{2} + 5 A D^{2} . . . (1)$

Now using pythagorus theorem,

In $△ A E B$ ,

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

In $△ A B C$ ,

$A C^{2} = A B^{2} + B C^{2} = y^{2} + 9 x^{2}$

In $△ A D B$ ,

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

Put value of $A E^{2}, A C^{2}$ & $A D^{2}$ in equation (1) we get,

LHS: $8 (4 x^{2} + y^{2}) = 32 x^{2} + 8 y^{2}$

RHS: $3 (y^{2} + 9 x^{2}) + 5 (y^{2} + x^{2}) = 27 x^{2} + 3 y^{2} + 5 x^{2} + 5 y^{2} = 32 x^{2} + 8 y^{2}$

LHS = RHS

Hence proved

Suggest Corrections

Similar questions

A B C

B . D

E

B C

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

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} .

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

Join BYJU'S Learning Program