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Maths-

General

Easy

Question

$AD comma BE comma CF text are the altitudes of a triangle end text ABC text . Prove that end text BD squared plus EC squared plus AF squared equals DC squared plus EA squared plus FB squared$

hint Hint:

Hint :- Applying pythagoras theorem to both triangles ABD and ADC. find the AD2 from both equations .Equate both the equations to get the required condition .
Get similarly equations for every altitude and add them

The correct answer is: Hence proved

Aim :- Prove that $B D squared plus E C squared plus A F squared equals D C squared plus E A squared plus F B squared$
Explanation(proof ) :-
Applying pythagoras theorem in ΔACD ,We get
$A C squared equals A D squared plus C D squared not stretchy rightwards double arrow A D squared equals A C squared minus C D squared$ – Eq1
Applying pythagoras theorem in ΔABD ,We get
$A B squared equals B D squared plus A D squared not stretchy rightwards double arrow A D squared equals A B squared minus B D squared$ —-Eq2
Substitute Eq2 in Eq1 ,
$A B squared minus B D squared equals A C squared minus C D squared not stretchy rightwards double arrow A B squared minus A C squared equals B D squared minus D C squared$ —- Eq3
Similarly , we get
$A C squared minus B C squared equals A F squared minus F B squared$ —- Eq4
$B C squared minus A B squared equals E C squared minus E A squared$ —- Eq5
Adding Eq3 ,Eq4 and Eq5
$table attributes columnspacing 1em end attributes row cell text We get, end text not stretchy rightwards double arrow A B squared minus A C squared plus A C squared minus B C squared plus B C squared minus A B squared equals B D squared minus D C squared plus A F squared minus F B squared plus E C squared minus E A squared end cell row blank row blank end table$
$not stretchy rightwards double arrow B D squared minus D C squared plus A F squared minus F B squared plus E C squared minus E A squared equals 0$
$not stretchy rightwards double arrow B D squared plus E C squared plus A F squared equals D C squared plus E A squared plus F B squared$
Hence proved

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parallel

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