If BL and CM are medians of a triangle ABC right angled at A, then prove that 4( BL2 + - Maths - Triangles

Question

If BL and CM are medians of a triangle ABC
right angled at A, then prove that 4( BL2 +
CM2 ) = 5 BC2

Priyanka Kedia · Accepted Answer

Given that: ABC be a right triangle right angled at A BL and CM are the medians of triangle ABC $T o \Pr o o v e : 4 ({B L}^{2} + {C M}^{2}) = 5 {B C}^{2}$

$\Pr o o f : I n Δ A B C, Δ A B L a n d Δ A C M, u \sin g P y t h a g o r a s t h e o r e m, w e h a v e, {B C}^{2} = {A C}^{2} + {A B}^{2} (1) A n d, {B L}^{2} = {A L}^{2} + {A B}^{2} \Rightarrow {B L}^{2} = {(\frac{A C}{2})}^{2} + {A B}^{2} (L i s t h e m i d p o int o f A C) \Rightarrow {B L}^{2} = \frac{{A C}^{2}}{4} + {A B}^{2} \Rightarrow 4 {B L}^{2} = {A C}^{2} + 4 {A B}^{2} (2) A n d, {C M}^{2} = {A C}^{2} + {A M}^{2} \Rightarrow {C M}^{2} = {(\frac{A B}{2})}^{2} + {A C}^{2} (M i s t h e m i d p o int o f A B) \Rightarrow {C M}^{2} = \frac{{A B}^{2}}{4} + {A C}^{2} \Rightarrow 4 {C M}^{2} = {A B}^{2} + 4 {A C}^{2} (3) A d d i n g e q u a t i o n (2) a n d (3), w e h a v e, 4 {B L}^{2} + 4 {C M}^{2} = {A C}^{2} + 4 {A B}^{2} + {A B}^{2} + 4 {A C}^{2} \Rightarrow 4 {B L}^{2} + 4 {C M}^{2} = 5 {A C}^{2} + 5 {A B}^{2} \Rightarrow 4 {B L}^{2} + 4 {C M}^{2} = 5 ({A C}^{2} + {A B}^{2}) \Rightarrow 4 {B L}^{2} + 4 {C M}^{2} = 5 {B C}^{2} (f r o m e q u a t i o n (1)) H e n c e p r o v e d$

If BL and CM are medians of a triangle ABC right angled at A, then prove that 4( BL2 + CM2 ) = 5 BC2

If BL and CM are medians of a triangle ABC right angled at A, then prove that 4( BL² + CM² ) = 5 BC²