Question

2
select the correct answer from each drop - down menu.
given: rhombus (abcd) with diagonal (overline{ac})
prove: (overline{ac}) bisects (angle dab) and (angle bcd)
identify the missing statement and reason in the proof.
given rhombus (abcd) with diagonal (overline{ac}), it follows from the definition of a rhombus that (overline{ab}congoverline{bc}congoverline{cd}congoverline{ad}). by the reflexive property of congruence, (overline{ac}congoverline{ac}). so, (\triangle daccong\triangle bac) by the (\boldsymbol{downarrow}). since corresponding parts of congruent triangles are congruent, (angle daccongangle\boldsymbol{downarrow}) and (angle dcacongangle\boldsymbol{downarrow}). by the definition of segment bisector (overline{ac}) bisects (angle dab) and (angle bcd).
drop - down menu options for the congruence criterion: aas theorem, asa criterion, sss criterion, sas criterion
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Explanation:

Step1: Recall Rhombus Properties

In a rhombus, all sides are congruent, so \( \overline{AB} \cong \overline{BC} \cong \overline{CD} \cong \overline{AD} \). The diagonal \( \overline{AC} \) is common to \( \triangle DAC \) and \( \triangle BAC \), so \( \overline{AC} \cong \overline{AC} \) (reflexive property).

Step2: Identify Congruence Criterion

We have three pairs of congruent sides: \( \overline{AD} \cong \overline{AB} \), \( \overline{CD} \cong \overline{BC} \), and \( \overline{AC} \cong \overline{AC} \). The SSS (Side - Side - Side) criterion states that if three sides of one triangle are congruent to three sides of another triangle, the triangles are congruent. So \( \triangle DAC \cong \triangle BAC \) by the SSS criterion.

Step3: Corresponding Angles

Since \( \triangle DAC \cong \triangle BAC \), their corresponding angles are congruent. So \( \angle DAC \cong \angle BAC \) and \( \angle DCA \cong \angle BCA \). This shows that \( \overline{AC} \) bisects \( \angle DAB \) and \( \angle BCD \).

Answer:

For the congruence criterion: SSS criterion
For the angle congruence: \( \angle BAC \) (for \( \angle DAC \cong \angle \square \)) and \( \angle BCA \) (for \( \angle DCA \cong \angle \square \))