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Equal Angles Subtend Equal Sides

BL and CM a...

Question

$B L$ and $C M$ are medians of $Δ A B C$ right angled at $A$ . Prove that $4 (B L^{2} + C M^{2}) = 5 B C^{2}$ .

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Solution

$G i v e n : △ A B C i s r i g h t a n g l e d a t A . B L a n d C M a r e i t s m e d i a n s T o P r o v e : P r o o f : I n △ B A L {B L}^{2} = {A L}^{2} + {A B}^{2} . . . (1) [U s i n g P y t h a g o r a s t h e o r e m] I n △ C A M {C M}^{2} = {A M}^{2} + {A C}^{2} . . . . . (2) [U s i n g P y t h a g o r a s t h e o r e m] A d d i n g (1) a n d (2) a n d t h e n m u l t i p l y i n g b y 4, w e g e t 4 ({B L}^{2} + {C M}^{2}) = 4 ({A L}^{2} + {A B}^{2} + {A M}^{2} + {A C}^{2}) = 4 {A L}^{2} + {A M}^{2} + ({A B}^{2} + {A C}^{2}) = 4 {A L}^{2} + {A M}^{2} + {B C}^{2} [I n r i g h t a n g l e A B C {A B}^{2} + {A C}^{2} = {B C}^{2}] = 4 ({A L}^{2} + {A M}^{2} + {B C}^{2}) = 4 ({M L}^{2} + {B C}^{2}) [I n L A M {A L}^{2} + {A M}^{2} = {M L}^{2}] = 4 {M L}^{2} + 4 {B C}^{2} [A c c o r d i n g t o m i d p o i n t t h e o r m a l i n e j o i n i n g m i d - p o i n t s o f t w o s i d e s i s p a r a l l e l t o t h i r d s i d e a n d i s e q u a l t o h a l f o f i t, M L = \frac{B C}{2}] = {B C}^{2} + {4 B C}^{2} = 5 {B C}^{2} . H e n c e p r o v e d$ .\quad

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4 (B L^{2} + C M^{2}) = 5 B C^{2}

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Q. BL and CM are medians of triangle ABC right angled at A , prove

4 (B L^{2} + C M^{2}) = 5 B C^{2}

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$4 ({B L}^{2} + {C M}^{2}) = 5 {B C}^{2}$ . [3 MARKS]

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