In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:
4(BL²+ CM²) = 5 BC²

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Solution

According to Pythagoras theorem,
In ∆ABC

${AB}^{2} + {AC}^{2} = {CB}^{2} . . . (1)$

In ∆ABC, point M is the midpoint of side BD.

$BM = MA = \frac{1}{2} AB$

${AC}^{2} + {BC}^{2} = 2 {CM}^{2} + 2 {AM}^{2} (by Apollonius theorem) . . . (2)$

In ∆CBA, point L is the midpoint of side AC.

$CL = LA = \frac{1}{2} AC$

${CB}^{2} + {AB}^{2} = 2 {BL}^{2} + 2 {AL}^{2} (by Apollonius theorem) . . . (3)$

Adding (2) and (3), we get

${AC}^{2} + {BC}^{2} + {CB}^{2} + {AB}^{2} = 2 {CM}^{2} + 2 {AM}^{2} + 2 {BL}^{2} + 2 {AL}^{2} \Rightarrow 2 {CM}^{2} + 2 {BL}^{2} = {AC}^{2} + {BC}^{2} + {CB}^{2} + {AB}^{2} - 2 {AM}^{2} - 2 {AL}^{2} \Rightarrow 2 ({CM}^{2} + {BL}^{2}) = {AC}^{2} + 2 {BC}^{2} + {AB}^{2} - 2 {AM}^{2} - 2 {AL}^{2} \Rightarrow 2 ({CM}^{2} + {BL}^{2}) = ({AC}^{2} + {AB}^{2}) + 2 {BC}^{2} - 2 {(\frac{1}{2} AB)}^{2} - 2 {(\frac{1}{2} AC)}^{2} \Rightarrow 2 ({CM}^{2} + {BL}^{2}) = {BC}^{2} + 2 {BC}^{2} - \frac{1}{2} {AB}^{2} - \frac{1}{2} {AC}^{2} (From (1)) \Rightarrow 2 ({CM}^{2} + {BL}^{2}) = 3 {BC}^{2} - \frac{1}{2} ({AB}^{2} + {AC}^{2}) \Rightarrow 2 ({CM}^{2} + {BL}^{2}) = 3 {BC}^{2} - \frac{1}{2} ({BC}^{2}) \Rightarrow 4 ({CM}^{2} + {BL}^{2}) = 2 [3 {BC}^{2} - \frac{1}{2} ({BC}^{2})] \Rightarrow 4 ({CM}^{2} + {BL}^{2}) = 6 {BC}^{2} - {BC}^{2} \Rightarrow 4 ({CM}^{2} + {BL}^{2}) = 5 {BC}^{2}$

Hence, 4(BL²+ CM²) = 5 BC².

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