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In Figure, AB...

Question

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

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Solution

Using Pythagoras Theorem in $Δ A D B$ and $Δ A D C$
In $Δ A D B$
On applying Pythagoras theorem,
$A B^{2} = A D^{2} + D B^{2} . . . . . (1)$
In $Δ A D C$
On applying Pythagoras therom,
$A C^{2} = A D^{2} + D C^{2}$
$A C^{2} = A D^{2} (D B + B C)^{2}$
$A C^{2} = A D^{2} + D B^{2} + B C^{2} + 2 (B C) (D B)$
From (1)
$\Rightarrow A C^{2} = A B^{2} + B C^{2} + 2 (B C) (D B)$
Hence Proved.

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