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Maths-

General

Easy

Question

$text In Triangle end text A B C comma straight angle B equals 90 to the power of ring operator text and end text D text is a point on end text B C text . Prove that end text A D squared plus B C squared equals A C squared plus B D squared$

hint Hint:

Hint :- Using pythagoras theorem in triangle ABC and ADB we find AC2 and AD2 and substitute the value of AD2 In AC2 equation to get the required result.

The correct answer is: Hence proved

Aim :- Prove that $A D squared plus B C squared equals A C squared plus B D squared$
Explanation(proof ) :-
As AP is median we get PB = $1 half$ BC also PB²= $1 fourth$ BC²
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 $A P squared equals A B squared plus P B squared text and end text P B squared equals 1 fourth B C squared text in the above equation end text$
$text We get end text A C squared equals open parentheses A P squared close parentheses plus B C squared minus 1 fourth B C squared not stretchy rightwards double arrow A C squared equals A P squared plus 3 over 4 B C squared$
$not stretchy rightwards double arrow A P squared equals A C squared minus 3 over 4 B C squared$
Hence proved

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