In fig., $A B C$ is a triangle in which $∠ A B C < 90^{o}$ and $A D ⊥ B C .$ Prove that $A C^{2} = A B^{2} + B C^{2} - 2 B C . B D$ .

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Solution

Proof:
In $△ A D C$
By Pythagoras Theorem,
$A C^{2} = A D^{2} + D C^{2} . . . . . . (1)$
In $△ A B D$
By Pythagoras Theorem,
$A B^{2} = A D^{2} + B D^{2} . . . . . . (2)$
Subtracting 1) and 2) we get,
$A C^{2} - A B^{2} = D C^{2} - B D^{2}$
$∴$ $A C^{2} - A B^{2} = D C^{2} - (B C - D C)^{2}$
$∴$ $A C^{2} - A B^{2} = 2 D C . B C - B C^{2}$
$∴$ $A C^{2} - A B^{2} = 2 (B C - B D) B C - B C^{2}$
$∴$ $A C^{2} - A B^{2} = - 2 D B . B C + 2 B C^{2} - B C^{2}$
$∴$ $A C^{2} = A B^{2} + B C^{2} - 2 B C . B D$

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In fig., $A B C$ is a triangle in which $∠ A B C > 90^{o}$ and $A D ⊥ C B$ produced. Prove that $A C^{2} = A B^{2} + B C^{2} + 2 B C . B D$ .

In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² − 2BC.BD.

In the given figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2BC.BD.

∠ B

△ A B C

A D ⊥ B C

{A C}^{2} - {A B}^{2} - {B C}^{2} + 2 B C . B D

△ A B C, m ∠ B = 90^{o}

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

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