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Maths-

General

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Question

$text In a Triangle end text A B C comma less than A equals 90 to the power of ring operator text and end text A D text is drawn perpendicular to end text B C text . Prove that end text A B squared plus C D squared equals B D squared plus A C squared$

hint Hint:

Hint :- using pythagoras theorem Find AC2 and AB2 subtract both the equation and substitute given conditions

The correct answer is: Hence proved

Aim :- Prove that $A B squared plus C D squared equals B D squared plus A C squared$
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 , We get

$A B squared minus B D squared equals A C squared minus C D squared$
$not stretchy left right double arrow A B squared plus C D squared equals B D squared plus A C squared$
Hence proved

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