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Maths-

General

Easy

Question

$text In Triangle end text A B C text , If end text A D text is the median show that, end text A B squared plus A C squared equals 2 open parentheses A D squared plus B D squared close parentheses$

hint Hint:

Hint :- Using pythagoras theorem in triangle ABE and ACE find AB2 and AC2
Now add the and using the midpoint condition and do necessary modification to get the required result.

The correct answer is: Hence proved

Aim :- Prove that $A C squared plus A B squared equals 2 open parentheses A D squared plus B D squared close parentheses$
Explanation(proof ) :-
AD is median so we get BD = DC
EC = DC-DE EC = BD - DE
Applying pythagoras theorem in ΔABD ,We get
$A B squared equals B E squared plus A E squared$
Substitute BE = BD+DE (from diagram)
$not stretchy rightwards double arrow A B squared equals left parenthesis B D plus D E right parenthesis squared plus A E squared$ —Eq1
Applying pythagoras theorem in ΔABD ,We get
$A C squared equals E C squared plus A E squared$
Substitute EC = BD - DE as we got above
$not stretchy rightwards double arrow A C squared equals left parenthesis B D minus D E right parenthesis squared plus A E squared$ —Eq2
Applying pythagoras theorem in ΔADE ,We get
$A D squared equals A E squared plus D E squared$ —Eq3
Adding Eq1 and Eq2 we get,
$open A B squared plus A C squared equals left parenthesis B D plus D E right parenthesis squared plus A E squared plus B D minus D E close parentheses squared plus A E squared$

As we know $left parenthesis a minus b right parenthesis squared plus left parenthesis a plus b right parenthesis squared equals 2 a squared plus 2 b squared$
$A B squared plus A C squared equals 2 B D squared plus 2 D E squared plus 2 A E squared equals 2 open parentheses B D squared plus open parentheses D E squared plus A E squared close parentheses close parentheses$
Substitute Eq3 in the above equation
We get, $A B squared plus A C squared equals 2 open parentheses B D squared plus A D squared close parentheses$
Hence proved

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