1. The problem asks for one additional fact that guarantees a frame with opposite sides congruent is rectangular. 2. A rectangle is a parallelogram with four right angles. 3. Since opposite sides are congruent, the frame is a parallelogram. 4. To guarantee it is a rectangle, one additional fact is that one angle is a right angle ($90^\circ$). 5. This is because in a parallelogram, if one angle is $90^\circ$, all angles are right angles, making it a rectangle. --- 1. Given quadrilateral ABCD is a rectangle with diagonals AC and BD intersecting at E. 2. We know in a rectangle, diagonals are congruent and bisect each other. 3. Given $AC = 3x - 11$ and $BE = x + 5$. 4. Since E is the midpoint of BD, $BE = ED = \frac{BD}{2}$. 5. Also, $AC = BD$ because diagonals of a rectangle are equal. 6. So, $BD = 2 \times BE = 2(x + 5) = 2x + 10$. 7. Set $AC = BD$: $$3x - 11 = 2x + 10$$ 8. Subtract $2x$ from both sides: $$3x - 11 - \cancel{2x} = \cancel{2x} + 10 - 2x$$ $$x - 11 = 10$$ 9. Add 11 to both sides: $$x - 11 + 11 = 10 + 11$$ $$x = 21$$ 10. Find $BD$: $$BD = 2x + 10 = 2(21) + 10 = 42 + 10 = 52$$ 11. The correct answer is B. 52.