Question

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 = dk	$square$	$square$
ck = ak	$square$	$square$
ck = dk	$square$	$square$

Explanation:

To determine if each condition is needed to prove \(ABCD\) is a parallelogram (using the theorem: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram), we analyze each:

1. \(BK = AK\)

For diagonals to bisect each other, we need \(AK = CK\) (bisect \(AC\)) and \(BK = DK\) (bisect \(BD\)). \(BK = AK\) does not relate to bisecting the diagonals (it involves segments of different diagonals). Thus, not needed.

2. \(BK = DK\)

This means diagonal \(BD\) is bisected at \(K\) (since \(BK = DK\)). A key part of the "diagonals bisect each other" theorem. Thus, needed.

3. \(CK = AK\)

This means diagonal \(AC\) is bisected at \(K\) (since \(CK = AK\)). A key part of the "diagonals bisect each other" theorem. Thus, needed.

4. \(CK = DK\)

\(CK\) (from \(AC\)) and \(DK\) (from \(BD\)) are segments of different diagonals. This does not relate to bisecting the diagonals. Thus, not needed.

Final Answers (Checkboxes):

• \(BK = AK\): \(\boldsymbol{\text{No}}\)
• \(BK = DK\): \(\boldsymbol{\text{Yes}}\)
• \(CK = AK\): \(\boldsymbol{\text{Yes}}\)
• \(CK = DK\): \(\boldsymbol{\text{No}}\)

Answer:

To determine if each condition is needed to prove \(ABCD\) is a parallelogram (using the theorem: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram), we analyze each:

1. \(BK = AK\)

For diagonals to bisect each other, we need \(AK = CK\) (bisect \(AC\)) and \(BK = DK\) (bisect \(BD\)). \(BK = AK\) does not relate to bisecting the diagonals (it involves segments of different diagonals). Thus, not needed.

2. \(BK = DK\)

This means diagonal \(BD\) is bisected at \(K\) (since \(BK = DK\)). A key part of the "diagonals bisect each other" theorem. Thus, needed.

3. \(CK = AK\)

This means diagonal \(AC\) is bisected at \(K\) (since \(CK = AK\)). A key part of the "diagonals bisect each other" theorem. Thus, needed.

4. \(CK = DK\)

\(CK\) (from \(AC\)) and \(DK\) (from \(BD\)) are segments of different diagonals. This does not relate to bisecting the diagonals. Thus, not needed.

Final Answers (Checkboxes):

• \(BK = AK\): \(\boldsymbol{\text{No}}\)
• \(BK = DK\): \(\boldsymbol{\text{Yes}}\)
• \(CK = AK\): \(\boldsymbol{\text{Yes}}\)
• \(CK = DK\): \(\boldsymbol{\text{No}}\)