If $A D ⊥ B C$ , prove that $A B^{2} + C D^{2} = B D^{2} + A C^{2}$

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Solution

In , $△ A D B, ∠ A D B = 90^{\circ}$

$⟹ A B^{2} = B D^{2} + A D^{2}$ (by Pythagorean theorem)..................(i)

$△ A D C, ∠ A D C = 90^{\circ}$

$⟹ A C^{2} = C D^{2} + A D^{2}$ (by Pythagorean theorem)..................(ii)

$(i) - (i i) ⟹ A B^{2} - A C^{2} = B D^{2} + A D^{2} - C D^{2} - A D^{2}$

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

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

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