Question
In a rectangle $\text{ABCD}$,prove that: $AC^2+ BD^2= AB^2+ BC^2+ CD^2+ DA^2$.
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Answer
Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Since, $\text{ABCD}$ is a rectangle angles $A, B, C$ and $D$ are rt. angles.
First, we consider the $\triangle A C D$, and applying Pythagoras theorem we get,
$AC ^2= DA ^2+ CD ^2\dots ....(i)$
Similarly, we get from rt. angle$ \triangle BDC$ we get,
$BD ^2= BC ^2+ CD ^2$
$=B C^2+A B^2\ldots .[$ In a rectangle, opposite sides are equal, $\therefore C D=AB ] \ldots (ii)$
Adding $(i)$ and $(ii)$
$A C^2+B D^2=A B^2+B C^2+C D^2+D A^2$
Hence proved.
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