The distances from a point to the corners of a rectangle are $6$, $7$, $9$, and (integer) $d$. Find $d$. 
Christina is standing in a rectangular garden. Her distances from the corners of the garden are $6$ meters, $7$ meters, $9$ meters, and $d$ meters, where $d$ is an integer. How to find $d$? 

Can someone lend me your hand on it?
 A: HINT
We have 5 unknown to determine, then we can guess at first $6$ opposite to $d$

with the following conditions


*

*$d^2-y^2=49-x^2$

*$49-w^2=36-z^2$

*$36-x^2=81-y^2$

*$81-z^2=d^2-w^2$
and then trying others configurations.
A: Let $P$ be a point inside a rectangle $XYZW$ such that $\{PX,PY,PZ\}=\{6,7,9\}$ and $PW=d$ is an integer.  Suppose that the projections of $P$ onto $XY$, $YZ$, $ZW$, and $WX$ are $A$, $B$, $C$, and $D$, respectively.  Write
$$x:=XA=WC\,,\,\,y:=YA=ZC\,,\,\,z:=YB=XD\,,\text{ and }w:=ZB=WD\,.$$
Therefore,
$$PX^2=x^2+z^2\,,\,\,PY^2=y^2+z^2\,,\text{ and }PZ^2=y^2+w^2\,.$$
This gives
$$d^2=PW^2=x^2+w^2=(x^2+z^2)+(y^2+w^2)-(y^2+z^2)=PX^2+PZ^2-PY^2\,.$$
The only possible choices for $(PX,PY,PZ)$ such that $d$ is an integer are $$(PX,PY,PZ)=(6,9,7)\text{ or }(PX,PY,PZ)=(7,9,6)\,,$$
for which $d=2$.  Below is an example of a possible configuration.  

A: We have a relation using Stewart's theorem for an arbitrary  parallelogram as follows "if ABCD is a parallelogram and P be any point in plane then $2AM^2 +2CM^2 - 2BM^2 + 2DM^2 $ = $ (AC^2 - BD^2)$ .
For a rectangle  AC =BD then by proceeding,  we can get desired un known. 
