Hello, Chocolatelover2!
You're misinterpreting the problem . . .
Prove: If two straight line segments bisect each other forming right angles,
any point on either of them is equal distance from the extremities of the other. Code:
* C

*P
*  *
*  *
*  *
A *    +    * B
O




* D
$\displaystyle AB$ and $\displaystyle CD$ are perpendicular bisectors of each other.
. . They intersect at $\displaystyle O.$
Hence: .$\displaystyle \angle COA = \angle COB = 90^o,\;OA = OB,\;OC = OD$
Let $\displaystyle P$ be any point on $\displaystyle CD.$
We want to prove that: $\displaystyle PA = PB.$
In $\displaystyle \Delta POA$ and $\displaystyle \Delta POB\!:\;\;\begin{array}{c}OA \:=\:OB \\ \angle POA \:=\:\angle POB \\ PO \:=\:PO\end{array}$
Hence: .$\displaystyle \Delta POA \cong \Delta POB$
Therefore: .$\displaystyle PA \:=\:PB$