BL&CM are medians of triangle ABC right angled at A. prove that 4(BL^2 +CM^2) =5BC^2.
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Given - A ∆ABC in which BL and CM are median of triangle and <A =90°
to proove-- 4(BL²+CM²) = 5 BC²
proof --In ∆ABC ,<A=90°
BC ² =AB ² +AC ²---1) [ by Pythagoras theorem ]
in ∆BAL ,<A =90°
=>BL ² = AL ² + AB ²
=>BL² = (1/2AC)² + AB² [ by Pythagoras theorem ]
=>4BL² = AC² +4AB²-----------2)
in ∆CAM ,<A =90°
•°• CM² =AM² +AC²
=>CM² = (1/2AB)² +AC²
=>CM² = (1/2AB)² + AC²
=>4CM² = AB² +4AC² --------------3)
now ,on adding 2) and 3 ) we get .
4(BL² + CM² )=5(AB²+AC²)
4(BL²+CL² )=5BC² ...from 1)
prooved. here ..
hope it helps you .
@Rajukumar111a
Given - A ∆ABC in which BL and CM are median of triangle and <A =90°
to proove-- 4(BL²+CM²) = 5 BC²
proof --In ∆ABC ,<A=90°
BC ² =AB ² +AC ²---1) [ by Pythagoras theorem ]
in ∆BAL ,<A =90°
=>BL ² = AL ² + AB ²
=>BL² = (1/2AC)² + AB² [ by Pythagoras theorem ]
=>4BL² = AC² +4AB²-----------2)
in ∆CAM ,<A =90°
•°• CM² =AM² +AC²
=>CM² = (1/2AB)² +AC²
=>CM² = (1/2AB)² + AC²
=>4CM² = AB² +4AC² --------------3)
now ,on adding 2) and 3 ) we get .
4(BL² + CM² )=5(AB²+AC²)
4(BL²+CL² )=5BC² ...from 1)
prooved. here ..
hope it helps you .
@Rajukumar111a
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