Question

prove: $\triangle abc cong \triangle cda$.step 1statement$angle b cong angle dLXB0overline{square} cong overline{square}$reasongivenselect a reason...search for reasons...reflexive propertytransitive propertysubstitutionin a triangle, sides opposite of congruent angles are congruenta segment bisector divides a segment into two congruent segmentsa midpoint divides a segment into two congruent segments

Explanation:

Step1: Identify alternate interior angles

Since $\overline{BC} \parallel \overline{AD}$, and $\overline{AC}$ is a transversal, $\angle BCA \cong \angle DAC$ (alternate interior angles theorem).

Step2: Identify common side

$\overline{AC} \cong \overline{CA}$ by Reflexive Property.

Step3: List given congruent angles

$\angle B \cong \angle D$ (Given).

Step4: Apply AAS congruence

We have two pairs of congruent angles and a non-included congruent side, so $\triangle ABC \cong \triangle CDA$ by AAS (Angle-Angle-Side) congruence.
For the missing statement in Step 1:

Step1: State common congruent side

$\overline{AC} \cong \overline{CA}$

Answer:

The missing statement is $\boldsymbol{\overline{AC} \cong \overline{CA}}$, and the completed proof shows $\triangle ABC \cong \triangle CDA$ by AAS congruence.