Complete Step 5 of the Proof

| Statement | Reason |
|------------------------------------------------|------------------------------------------------|
| 1. Quadrilateral [tex]\(ABCD\)[/tex], with [tex]\(A(1,3), B(2,4), C(4,2), D(3,1)\)[/tex] | Given |
| 2. Draw [tex]\(\overline{AC}\)[/tex] and [tex]\(\overline{BD}\)[/tex] | A line segment can be drawn between any two points |
| 3. [tex]\(AC = \sqrt{(4-1)^2+(2-3)^2} = \sqrt{9+1} \approx 3.2\)[/tex] | Distance formula |
| [tex]\(BD = \sqrt{(3-1)^2+(1-4)^2} = \sqrt{8+1} \approx 3.2\)[/tex] | |
| 4. [tex]\(AC = BD\)[/tex] | Definition of equality |
| 5. Quadrilateral [tex]\(ABCD\)[/tex] is a rectangle. | |

What is the statement for Step 5?



Answer :

To complete Step 5 of the proof, we need to establish that the diagonals of the quadrilateral are actually equal and use this information to state that Quadrilateral [tex]\(ABCD\)[/tex] is a rectangle, supported by the conclusion from the previous steps.

Here’s the detailed, step-by-step completion of the proof, including the statement and reason for Step 5:

```
Statement Reason
1. Quadrilateral [tex]\(ABCD\)[/tex], with [tex]\(A(1,3)\)[/tex], [tex]\(B(2,4)\)[/tex], Given
[tex]\(C(4,2)\)[/tex], [tex]\(D(3,1)\)[/tex]

2. Draw [tex]\(\overline{AC}\)[/tex] and [tex]\(\overline{BD}\)[/tex]. A line segment can be drawn between any two points.

3. [tex]\(AC = \sqrt{(4-1)^2 + (2-3)^2} Distance formula = \sqrt{9 + 1} \approx 3.2\)[/tex]

[tex]\(BD = \sqrt{(3-1)^2 + (1-4)^2} = \sqrt{8 + 1} \approx 3.2\)[/tex]

4. [tex]\(AC = BD\)[/tex] Definition of equality

5. Quadrilateral [tex]\(ABCD\)[/tex] is a rectangle. Since the lengths of the diagonals are equal, [tex]\(AC = BD\)[/tex],
and in this specific arrangement of points, this condition implies
that the quadrilateral is a rectangle.
```

Thus, the statement for Step 5 is:
"Quadrilateral [tex]\(ABCD\)[/tex] is a rectangle."

The reason for Step 5 is based on the properties of a rectangle: in a rectangle, the diagonals are congruent (equal in length). Given that the lengths of the diagonals [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] are equal and applying this condition, we can conclude that Quadrilateral [tex]\(ABCD\)[/tex] is a rectangle.