Question
$O$ is any point inside a rectangle $\text{ABCD}$.Prove that: $OB^2+ OD^2= OC^2+ OA^2$.
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Answer
Draw rectangle $\text{ABCD}$ with arbitrary point $O$ within it, and then draw lines $O A, O B, O C, O D$. Then draw lines from point $O$ perpendicular to the sides: $OE , OF , OG , OH$.
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.
Using Pythagorean theorem we have from the above diagram:
$ O A^2=A H^2+O H^2=A H^2+A^2$
$ O C^2=C G^2+O G^2= EB ^2+ HD ^2$
$ O B^2=E O^2+ BE ^2=A H^2+ BE ^2$
$ O D^2= HD ^2+ OH ^2= HD ^2+ AE ^2$
Adding these equalities we get:
$O A^2+O C^2=A H^2+H D^2+A E^2+E B^2$
$ O B^2+O D^2=A H^2+H D^2+A E^2+E B^2$
From which we prove that for any point within the rectangle there is the relation
$O A^2+O C^2=O B^2+O D^2$
Hence Proved.
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