In a triangle ABC, O is …

In right triangles {tex}\triangle{/tex}ODB and {tex}\triangle{/tex}ODC, we have
OB² = OD² + BD²  [ Using Pythagoras Theorem] 
and, OC² = OD² + CD^{2
​​​​}Subtracting them, we get
{tex}\therefore{/tex} OB² - OC² = (OD² + BD²) - (OD² + CD²)
{tex}\Rightarrow{/tex} OB² - OC² = BD² - CD² ...(i)
Similarly, we have
OC² - OA² = CE² - AE² ....(ii) [ Using Pythagoras Theorem In {tex}\triangle {/tex}AOE and {tex}\triangle {/tex}COE] 
and, OA² - OB² = AF² - BF² ...(iii) [ Using Pythagoras Theorem In {tex}\triangle{/tex} AOF and {tex}\triangle {/tex}BOF] 
Adding (i), (ii) and (iii), we get
(OB² - OC²) + (OC² - OA²) + OA² - OB²) = (BD² - CD²) + (CE² - AE²) +(AF² - BF²)
{tex}\Rightarrow{/tex} (BD² + CE² + AF²) - (AE² + CD² + BF²) = 0
{tex}\Rightarrow{/tex} AF² + BD² + CE² = AE² + BF² + CD²
​​​​​Hence Proved