the diagonals of quadrilateral abcd intersect at point k. is each of the following needed to prove that abcd is a parallelogram?
| | yes | no |
| bk = ak | □ | □ |
| bk = dk | □ | □ |
| ck = ak | □ | □ |
| ck = dk | □ | □ |

Question

Accepted Answer

To determine if each condition is needed to prove $ABCD$ is a parallelogram, we use the theorem: *A quadrilateral is a parallelogram if its diagonals bisect each other* (i.e., the point of intersection $K$ divides each diagonal into two equal parts, so $AK = CK$ and $BK = DK$).

### 1. Analyzing $BK = AK$
The theorem requires diagonals to bisect each other ($AK = CK$, $BK = DK$), not $BK = AK$. This condition does not relate to the diagonals bisecting each other. Thus, $BK = AK$ is **not** needed.

### 2. Analyzing $BK = DK$
By the "diagonals bisect each other" theorem, if $BK = DK$ (and $AK = CK$), the diagonals bisect each other, proving $ABCD$ is a parallelogram. Thus, $BK = DK$ is **needed**.

### 3. Analyzing $CK = AK$
By the "diagonals bisect each other" theorem, if $CK = AK$ (and $BK = DK$), the diagonals bisect each other, proving $ABCD$ is a parallelogram. Thus, $CK = AK$ is **needed**.

### 4. Analyzing $CK = DK$
The theorem requires diagonals to bisect each other ($AK = CK$, $BK = DK$), not $CK = DK$. This condition does not relate to the diagonals bisecting each other. Thus, $CK = DK$ is **not** needed.

### Filling the Table:
| Condition | Yes | No |
|-----------|-----|----|
| $BK = AK$ | ☐ | ☑ |
| $BK = DK$ | ☑ | ☐ |
| $CK = AK$ | ☑ | ☐ |
| $CK = DK$ | ☐ | ☑ |

### Final Answers (Checked Boxes):
- $BK = AK$: No (check the "No" box).
- $BK = DK$: Yes (check the "Yes" box).
- $CK = AK$: Yes (check the "Yes" box).
- $CK = DK$: No (check the "No" box).

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QUESTION IMAGE

Question

Explanation:

1. Analyzing \(BK = AK\)

2. Analyzing \(BK = DK\)

3. Analyzing \(CK = AK\)

4. Analyzing \(CK = DK\)

Filling the Table:

Final Answers (Checked Boxes):

Answer: