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

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Solution

Given that $A B C$ is a triangle, right-angled at $A .$
Also, given that BL and CM are the medians.
To prove: $4 (B L^{2} + C M^{2}) = 5 B C^{2}$
Proof:
Now, consider the triangle $A B L$ ,
$B L^{2} = A B^{2} + A L^{2}$
$B L^{2} = A B^{2} + {(\frac{A C}{2})}^{2}$ [Since, BL is the median]
Now, take the triangle $A C M, C M^{2} = A C^{2} + A M^{2}$
$C M^{2} = A C^{2} + {(\frac{A B}{2})}^{2}$ [Since CM is the median]
Therefore, $B L^{2} + C M^{2} = A B^{2} + A C^{2} + {(\frac{A C}{2})}^{2} + {(\frac{A B}{2})}^{2}$
$B L^{2} + C M^{2} = A B^{2} + A C^{2} + (\frac{A C^{2}}{4}) + (\frac{A B^{2}}{4})$
$4 (B L^{2} + C M^{2}) = 5 A B^{2} + 5 A C^{2}$
$4 (B L^{2} + C M^{2}) = 5 (A B^{2} + A C^{2})$
$4 (B L^{2} + C M^{2}) = 5 B C^{2}$
Hence, proved.

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