In a parallelogram, the diagonals bisect each other. This means that each diagonal is divided into two equal segments at the point of intersection.
Let’s denote the intersection point of diagonals AC and BD as point O. Given that AC = 48, the segments AO and OC are each half of AC. Therefore:
\[ AO = OC = \frac{AC}{2} = \frac{48}{2} = 24 \]
Additionally, it is given that \(DO = 7x + 3\). Since diagonals bisect each other, \(DO\) is equal to \(BO\). Hence:
\[ BO = DO = 7x + 3 \]
In a rectangle, the diagonals are congruent. Therefore, we need to also have:
\[ AO = BO \]
Substituting the values we calculated:
\[ 24 = 7x + 3 \]
Now, we can solve for \(x\):
- Subtract 3 from both sides:
\[ 24 - 3 = 7x \]
\[ 21 = 7x \]
- Divide both sides by 7:
\[ x = \frac{21}{7} = 3 \]
Thus, the value of \(x\) that proves that parallelogram ABCD is a rectangle is:
\[ \boxed{3} \]