If AD perpendicular to BC , prove that AB² + CD² = BD² +AC²

1.	If AD perpendicular to BC , prove that AB² + CD² = BD² +AC²
Answer» To Prove :- $\leadsto\sf\: AB {}^{<klux>2</klux>} + CD {}^{2} = BD {}^{2} + AC {}^{2}$ SOLUTION:- From triangle ADC , we have $\leadsto \sf \: AC {}^{2} = AD {}^{2} + CD {}^{2} \longrightarrow(Pythagoras \: therom ) \longrightarrow(<klux>1</klux>)$ From triangle ADB , we have $\leadsto \sf \: AB {}^{2} = AD {}^{2} + BD {}^{2} \longrightarrow(Pythagoras \: therom ) \longrightarrow(2)$ Subtracting (1) from (2), we have $\leadsto \sf \: AB {}^{2} - AC {}^{2} = BD {}^{2} - CD {}^{2} \\ \\ \sf \: (or)\\ \\\leadsto \sf \: AB {}^{2} + CD {}^{2} = BD {}^{2} +AC {}^{2}$ Hence PROVED ...

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