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Maths-

General

Easy

Question

P and Q are points on the sides CA and CB respectively of a $straight triangle ABC$ right angled at c. Prove that $AQ squared plus BP squared equals$ $A B squared plus P Q squared$

The correct answer is: Hence proved .

Aim :- Prove that $A Q squared plus B P squared equals A B squared plus P Q squared$
Hint :- Applying pythagoras theorem to both triangles ABC, ACQ,PCQ and PCB
find the equation, add them and substitute proper conditions to prove the condition

Hint :- Applying pythagoras theorem to both triangles ABC, ACQ,PCQ and PCB
find the equation, add them and substitute proper conditions to prove the condition
Explanation(proof ) :-
Applying pythagoras theorem to triangle AQC we get ,
$A Q squared equals A C squared plus Q C squared minus Eq 1$
Applying pythagoras theorem to triangle PBC we get ,
$B P squared equals P C squared plus B C squared minus Eq 2$
Applying pythagoras theorem to triangle ABC we get ,
$A B squared equals A C squared plus B C squared minus Eq 3$
Applying pythagoras theorem to triangle PCQ we get ,
$P Q squared equals P C squared plus Q C squared minus Eq 4$
Adding Eq1 and Eq2 we get ,
$A Q squared plus B P squared equals A C squared plus Q C squared plus P C squared plus B C squared$
$A Q squared plus B P squared equals open parentheses A C squared plus B C squared close parentheses plus open parentheses Q C squared plus P C squared close parentheses left curly bracket text substitute Eq3 and Eq4} end text$
$A Q squared plus B P squared equals A B squared plus P Q squared$
Hence proved

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