$\begin{matrix} I n t h e f o l l o w i n g f i g u r e, A D i s p e r p e n d i c u l a r t o B C a n d D d i v i d e s B C i n t h e r a t i o 1 : 3 Prove t h a t : 2 A C^{2} = 2 A B^{2} + B C^{2} \end{matrix}$

Open in App

Solution

Given : $A D ⊥ B C$ and $\frac{B D}{D C} = \frac{1}{3}$
$B D + D C = B C$
$B D + 3 B D = B C ⟹ B D = \frac{B C}{4}$
Hence $D C = \frac{3 B C}{4}$
By pythagoras theorem in $△ A D C, A C^{2} = A D^{2} + D C^{2}$ ---(1)

In $△ A B D, A B^{2} = B D^{2} + A D^{2}$ ---(2)

Subtracting (1) from (2), $A C^{2} - A B^{2} = D C^{2} - B D^{2} = (D C + B D) (D C - B D) = B C (\frac{3 B C}{4} - \frac{B C}{4}) = \frac{B C^{2}}{2}$

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

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

Suggest Corrections

Similar questions

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

Δ A B C

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

△ A B C, s e g A D ⊥ s e g B C, D B = 3 C D

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

Δ

⊥

= 3

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

Join BYJU'S Learning Program