Question

deltamath
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triangle proofs (no cpctc)
no late credit
grade: 0%
basic triangle proofs (congruence only - no cpctc)
prove: $\triangle abc \cong \triangle cda$.
step 1
statement
$\angle b \cong \angle d$
$\overline{bc} \parallel \overline{ad}$
$\overline{\quad\quad} \cong \overline{\quad\quad}$
reason
given

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reflexive property
transitive property
substitution
in a triangle, sides opposite of congruent angles are congruent
a segment bisector divides a segment into two congruent segments
a 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 reflexive side

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

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 the Angle-Angle-Side (AAS) Congruence Theorem.

For the missing statement in Step 1:

Step1: Fill congruent alternate angles

$\angle BCA \cong \angle DAC$
Reason: Alternate Interior Angles Theorem (from parallel lines $\overline{BC} \parallel \overline{AD}$)

Answer:

The completed statement for Step 1 is $\boldsymbol{\angle BCA \cong \angle DAC}$ with the reason "Alternate Interior Angles Theorem". The final proof confirms $\boldsymbol{\triangle ABC \cong \triangle CDA}$ by AAS congruence.