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Maths-

General

Easy

Question

$A B C text is a right triangle with end text straight angle A equals 90 to the power of ring operator BL text and end text CM text are its medians. Prove that end text 4 open parentheses BL squared plus CM squared close parentheses equals 5 BC squared text . end text$

The correct answer is: Hence proved

Solution :-
Aim :- Prove that $4 open parentheses B L squared plus C M squared close parentheses equals 5 B C squared$
Hint :- Applying pythagoras theorem to both triangles ACM and BLA. find the BL²
and CM² add them and substitute proper conditions to prove the condition .
Explanation(proof ) :-
Given, BL and CM are the medians of a triangle ABC
M is midpoint of AB i.e AM = ½ AB
L is mid point of AC i.e AL = ½ AC
Applying pythagoras theorem to triangle BLA we get ,
$B L squared equals A L squared plus A B squared$
As we know AL = ½ AC we get $B L squared equals A B squared plus fraction numerator A C squared over denominator 4 end fraction minus Eq 1$
Applying pythagoras theorem to triangle ACM we get ,
$C M squared equals A C squared plus A M squared$
As we know AM = ½ AB we get $C M squared equals A C squared plus fraction numerator A B squared over denominator 4 end fraction minus Eq 2$ — Eq2
Applying pythagoras theorem to triangle ABC we get ,
$A B squared plus A C squared equals B C squared minus Eq 3$

Adding Eq1 and Eq2 we get ,
$B L squared plus C M squared equals A B squared plus fraction numerator A C squared over denominator 4 end fraction plus A C squared plus fraction numerator A B squared over denominator 4 end fraction$
$not stretchy rightwards double arrow B L squared plus C M squared equals 5 over 4 open parentheses A B squared plus A C squared close parentheses$
Substitute Eq 3 in the above , we get
$B L squared plus C M squared equals 5 over 4 open parentheses B C squared close parentheses not stretchy rightwards double arrow 4 open parentheses B L squared plus C M squared close parentheses equals 5 B C squared$
Hence proved

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