Question

In $ΔABC, ∠ BAC = 90 °$ $seg BL$ and $seg CM$ are medians of $△ ABC$ . Then prove that $4 ({BL}^{2} + {CM}^{2}) = 5 {BC}^{2}$

Accepted Answer

In a right-angled triangle on the hypotenuse is equal to sum of the squares on the other two sides.

Given:- $BL & CM$ are medians

Then $AL = \frac{AC}{2} & AM = \frac{AB}{2} . . . . . . (1)$

In $∆ BAL$ , by Pythagoras theorem

${BL}^{2} = {AB}^{2} + {AL}^{2} - (2)$

In $∆ MAC$ , by Pythagoras theorem

${Cm}^{2} = {Am}^{2} + {AC}^{2} . . . . (3)$

In $∆ BAC$ , by Pythagoras theorem

${BC}^{2} = {AB}^{2} + {AC}^{2} . . . . . . (4)$

Adding (2) & (3) we get

${BL}^{2} + {CD}^{2} = {AB}^{2} + {AL}^{2} + {AM}^{2} + {AC}^{2}$

${BL}^{2} + C M^{2} = ({AB}^{2} + {AC}^{2}) + {AL}^{2} + {AM}^{2}$

Using (4), we get

${BL}^{2} + C M^{2} = {BC}^{2} + {AL}^{2} + {AM}^{2}$

Using (1),we get

${BL}^{2} + {CM}^{2} = {BC}^{2} + (\frac{{AC}^{2}}{4}) + \frac{{AB}^{2}}{4}$

${BL}^{2} + {CM}^{2} = {BC}^{2} + \frac{1}{4} ({AB}^{2} + {RC}^{2})$

${BL}^{2} + C M^{2} = {BC}^{2} + \frac{1}{4} {BC}^{2}$

${BL}^{2} + C M^{2} = 4 {BC}^{2} + {BC}^{2}$

${BL}^{2} + {CM}^{2} = \frac{5 {BC}^{2}}{4}$

$4 ({BL}^{2} + {CM}^{2}) = 5 {BC}^{2}$

Hence proved.

Maharashtra Board Solutions for Chapter: Pythagoras Theorem, Exercise 3: Problem set 2

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Pythagoras Theorem, Exercise 3: Problem set 2

Questions from Maharashtra Board Solutions for Chapter: Pythagoras Theorem, Exercise 3: Problem set 2 with Hints & Solutions

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