In $△ A B C$ , AD is perpendicular to BC. Prove that : ${A B}^{2} + {C D}^{2} = {A C}^{2} + {B D}^{2}$

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Solution

Since triangles ABD and ACD are right-angled triangles at D.

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

and, ${A C}^{2} = {A D}^{2} + {C D}^{2}$ .....(ii)

Subtracting (ii) from (i), we get

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

$\Rightarrow$ ${A B}^{2} + {C D}^{2} = {A C}^{2} + {B D}^{2}$ $[H e n c e p r o v e d]$

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