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Maths-

General

Easy

Question

$text Prove that in a end text straight triangle A B C text , if end text A D text is perpendicular to end text B C text , then end text A B squared plus C D squared equals A C squared plus B D squared text . end text$

The correct answer is: Hence proved

Solution :-
Aim :- Prove that $A B squared plus C D squared equals A C squared plus B D squared$
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
Explanation(proof ) :-
Applying pythagoras theorem in ΔACD ,We get
$A C squared equals A D squared plus C D squared not stretchy rightwards double arrow A D squared equals A C squared minus C D squared minus Eq 1$
Applying pythagoras theorem in ΔABD ,We get
$A B squared equals B D squared plus A D squared not stretchy rightwards double arrow A D squared equals A B squared minus B D squared minus Eq 2$
Substitute Eq2 in Eq1 ,
$A B squared minus B D squared equals A C squared minus C D squared not stretchy rightwards double arrow A B squared plus C D squared equals A C squared plus B D squared$
$therefore A B squared plus C D squared equals A C squared plus B D squared$

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