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Distance between Two Points Using Pythagoras Theorem

In ABC,seg ...

Question

In $△ A B C, s e g A D ⊥ s e g B C, D B = 3 C D$ , then
$2 A B^{2} = 2 A C^{2} + B C^{2}$

True

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False

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Solution

The correct option is A True
Given that in $Δ A B C$ , we have

$A D ⊥ B C$ and $B D = 3 C D$

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

$A B^{2} = A D^{2} + B D^{2} . . . (i)$

$A C^{2} = A D^{2} + D C^{2} . . . (i i)$ [By Pythagoras theorem]

Subtracting equation $(i i)$ from equation $(i)$ , we get

$A B^{2} - A C^{2} = B D^{2} - D C^{2}$

$= 9 C D^{2} - C D^{2} [∴ B D = 3 C D]$

$= 8 C D^{2} = 8 (\frac{B C}{4})^{2}$ [Since, $B C = D B + C D = 3 C D + C D = 4 C D$ ]

Therefore, $A B^{2} - A C^{2} = \frac{B C^{2}}{2}$

$\Rightarrow 2 (A B^{2} - A C^{2}) = B C^{2}$

$\Rightarrow 2 A B^{2} - 2 A C^{2} = B C^{2}$

$∴ 2 A B^{2} = 2 A C^{2} + B C^{2}$ .

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Similar questions

The perpendicular from A on side BC of a ΔABC intersect BC at D such that DB = 3 CD. Prove that 2 AB² = 2 AC² + BC²

A D

B C

△ A B C

B C

D

D B = 3 C D

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

A D ⊥ B C

B D = 3 C D

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

A

B C

B C

D

D B = 3 C D

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

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