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Maths-

General

Easy

Question

$A B C D text is a rhombus, prove that end text A C squared plus B D squared equals 4 times A B squared$

hint Hint:

Hint :- Applying pythagoras theorem to both triangles AIB. find the AB2 and multiply by 4 and substitute proper conditions to prove the condition .

The correct answer is: Hence proved .

Aim :- Prove that $A C squared plus B D squared equals 4 times A B squared$
Explanation(proof ) :-
Given ABCD is a rhombus
The diagonals in the rhombus are perpendicular bisectors.
BI = $1 half$ BD — Eq 1
AI = $1 half$ AC — Eq 2
Applying pythagoras theorem in ΔAIB,
$A B squared equals A I squared plus B I squared$

Multiplying with 4 on both sides
$4 A B squared equals 4 A I squared plus 4 B I squared not stretchy rightwards double arrow 4 times A B squared equals left parenthesis 2 A I right parenthesis squared plus left parenthesis 2 B I right parenthesis squared$
Applying Eq 1 and 2
We get, $text 4. end text A B squared equals A C squared plus B D squared$
Hence proved .

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