Question

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

Answer 1

To Prove :-

$\leadsto\sf\: AB {}^{2} + CD {}^{2} = BD {}^{2} + AC {}^{2}$

Solution:-

• From triangle ADC , we have

$\leadsto \sf \: AC {}^{2} = AD {}^{2} + CD {}^{2} \longrightarrow(Pythagoras \: therom ) \longrightarrow(1)$

• From triangle ADB , we have

$\leadsto \sf \: AB {}^{2} = AD {}^{2} + BD {}^{2} \longrightarrow(Pythagoras \: therom ) \longrightarrow(2)$

• Subtracting (1) from (2), we have

$\leadsto \sf \: AB {}^{2} - AC {}^{2} = BD {}^{2} - CD {}^{2} \\ \\ \sf \: (or)\\ \\</p><p></p><p>\leadsto \sf \: AB {}^{2} + CD {}^{2} = BD {}^{2} +AC {}^{2}$

Hence Proved ...

Answer 2

Answer:

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