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Maths-

General

Easy

Question

The angle B of a triangle ABC is right angled triangle. AP is the median of the triangle. Prove that
$A P squared equals A C squared minus 3 over 4 B C squared$

hint Hint:

Hint :- Using pythagoras theorem in triangle ABC and APB we find AC2 and AP2 and substitute the value of AP2 In AC2 equation to get the required result.

The correct answer is: Hence proved

Aim :- Prove that $A P squared equals A C squared minus 3 over 4 B C squared$
Explanation(proof ) :-
As AP is median we get PB = $1 half$ BC also = $1 fourth B C squared$
Applying pythagoras theorem in triangle APB we get
$A P squared equals A B squared plus P B squared$
Applying pythagoras theorem in triangle ABC we get
$A C squared equals A B squared plus B C squared$

Adding and subtracting PB² we get , $A C squared equals open parentheses A B squared plus P B squared close parentheses plus B C squared minus P B squared$
Substitute $A P squared equals A B squared plus P B squared$ and $P B squared equals 1 fourth B C squared$ in the above equation
We get, $A C squared equals open parentheses A P squared close parentheses plus B C squared minus 1 fourth B C squared not stretchy rightwards double arrow A C squared equals A P squared plus 3 over 4 B C squared$
$not stretchy rightwards double arrow A P squared equals A C squared minus 3 over 4 B C squared$
Hence proved

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