TITLE: Given distances from a point to three vertices of a rectangle, find the distance to the fourth vertex
QUESTION [5 upvotes]: Problem: $ABCD$ is a rectangle. A point $P$ is $11$ from $A$, $13$ from $B$ and $7$ from $C$. What is the length of $DP=x?$ (Note: $P$ can be inside the rectangle or outside of it.)


I drew this scenario as best as I could, but I only have triangles with two sides and no angles. How do I begin? Any weird and relatively unknown theorems I should use?

REPLY [2 votes]: Just $x^2+13^2=7^2+11^2$ by Pythagoras theorem.
Let $PK$, $PL$, $PM$ and $PN$ be altitudes of $\Delta DPA$, $\Delta DPC$, $\Delta CPB$ and $\Delta APB$ respectively.
Thus,
$$x^2+13^2=DK^2+KP^2+PM^2+MB^2=PN^2+KP^2+PM^2+PL^2=11^2+7^2$$