In the adjoining figure $A D ⊥$ $B C, B - D - C$ show that ${A B}^{2} - {B D}^{2} = {A C}^{2} - {C D}^{2}$ .

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Solution

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From Pythagoras theorem,
$A D^{2} + B D^{2} = A B^{2} \Rightarrow A D^{2} = A B^{2} - B D^{2}$ (1)
$A D^{2} + C D^{2} = A C^{2} \Rightarrow A D^{2} = A C^{2} - C D^{2}$ (2)
From (1) & (2), $A B^{2} - B D^{2} = A C^{2} - C D^{2}$ .

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