warm-up #1 (2/5/26)
given:
what information is needed to prove the two triangles congruent using aas?

Question

Accepted Answer

# Explanation:
## Step1: Recall AAS Congruence
AAS (Angle - Angle - Side) congruence criterion states that if two angles and a non - included side of one triangle are congruent to the corresponding two angles and non - included side of another triangle, then the two triangles are congruent.

In the parallelogram $ABCD$ (implied by $AB\parallel DC$ and $AD\parallel BC$ from the arrows), we have $\triangle ADC$ and $\triangle CBA$. We know that $\angle DAC=\angle BCA$ (alternate interior angles, since $AD\parallel BC$ and $AC$ is a transversal) and $\angle DCA = \angle BAC$ (alternate interior angles, since $AB\parallel DC$ and $AC$ is a transversal). Also, $AC$ is a common side. But for AAS, we need two angles and a non - included side.

Looking at the triangles $\triangle ACD$ and $\triangle CAB$:
- We already have that $AB\parallel DC$ and $AD\parallel BC$ (from the arrows on the sides, indicating parallel sides, so it's a parallelogram). So, $\angle D=\angle B$ (opposite angles of a parallelogram are equal) and $\angle DAC=\angle BCA$ (alternate interior angles) or $\angle DCA=\angle BAC$ (alternate interior angles).
- To use AAS, we need either:
    - One pair of equal angles (other than the ones from parallel lines) and a pair of equal non - included sides. Or, since we know that $AC$ is common, if we know that $\angle D=\angle B$, $\angle DAC=\angle BCA$ and $AD = BC$ (or $\angle DCA=\angle BAC$ and $DC = AB$) or $\angle D=\angle B$, $\angle DCA=\angle BAC$ and $DC=AB$ (or $AD = BC$). But more precisely, in the triangles $\triangle ADC$ and $\triangle CBA$:
      - We know that $AC$ is a side common to both triangles. For AAS, we need two angles and a non - included side. Let's consider the angles:
        - Since $AB\parallel DC$ and $AD\parallel BC$, $\angle D$ and $\angle B$ are equal (opposite angles of parallelogram). Also, $\angle DAC=\angle BCA$ (alternate interior angles) and $\angle DCA=\angle BAC$ (alternate interior angles).
        - To apply AAS, we need either:
          - $\angle D=\angle B$, $\angle DAC=\angle BCA$ and $AD = BC$ (because in $\triangle ADC$, the angles $\angle D$ and $\angle DAC$ have non - included side $AD$, and in $\triangle CBA$, the angles $\angle B$ and $\angle BCA$ have non - included side $BC$).
          - Or $\angle D=\angle B$, $\angle DCA=\angle BAC$ and $DC = AB$ (because in $\triangle ADC$, the angles $\angle D$ and $\angle DCA$ have non - included side $DC$, and in $\triangle CBA$, the angles $\angle B$ and $\angle BAC$ have non - included side $AB$).

- Alternatively, since $AC$ is common, if we know that $\angle D=\angle B$, and either $\angle DAC=\angle BCA$ with $AD = BC$ or $\angle DCA=\angle BAC$ with $DC = AB$, we can use AAS. But the key information needed is: We need to know that two angles and a non - included side are congruent. In the context of the parallelogram (from the diagram with parallel sides), we know that $AB\parallel DC$ and $AD\parallel BC$, so $\angle D=\angle B$ (opposite angles of parallelogram). Then, if we know that $\angle DAC=\angle BCA$ and $AD = BC$ (or $\angle DCA=\angle BAC$ and $DC = AB$), we can use AAS. But more simply, since $AC$ is a common side, and we have $\angle D=\angle B$ (from parallelogram), we need one more pair of angles (like $\angle DAC=\angle BCA$ or $\angle DCA=\angle BAC$) and the non - included side ( $AD = BC$ or $DC = AB$). But the most direct way: In $\triangle ADC$ and $\triangle CBA$, we have $AC = AC$ (common side). For AAS, we need two angles and a non - included side. So we need to know that $\angle D=\angle B$, $\angle DAC=\angle BCA$ and $AD = BC$ (or $\angle D=\angle B$, $\angle DCA=\angle BAC$ and $DC = AB$). But since the figure is a parallelogram (from the parallel arrows on the sides), $AD = BC$ and $DC = AB$ (opposite sides of parallelogram are equal) and $\angle D=\angle B$ (opposite angles of parallelogram are equal). So if we take $\angle D=\angle B$, $\angle DAC=\angle BCA$ and $AD = BC$ (non - included side between $\angle D$ and $\angle DAC$ in $\triangle ADC$ and between $\angle B$ and $\angle BCA$ in $\triangle CBA$) or $\angle D=\angle B$, $\angle DCA=\angle BAC$ and $DC = AB$ (non - included side between $\angle D$ and $\angle DCA$ in $\triangle ADC$ and between $\angle B$ and $\angle BAC$ in $\triangle CBA$).

- Another way: The two triangles are $\triangle ACD$ and $\triangle CAB$. We know that $AC$ is common. To use AAS, we need two angles and a non - included side. So we need either:
        - $\angle D=\angle B$, $\angle DAC=\angle BCA$, and $AD = BC$ (because in $\triangle ACD$, $\angle D$ and $\angle DAC$ have side $AD$ as non - included, and in $\triangle CAB$, $\angle B$ and $\angle BCA$ have side $BC$ as non - included)
        - Or $\angle D=\angle B$, $\angle DCA=\angle BAC$, and $DC = AB$ (because in $\triangle ACD$, $\angle D$ and $\angle DCA$ have side $DC$ as non - included, and in $\triangle CAB$, $\angle B$ and $\angle BAC$ have side $AB$ as non - included)

- Since the figure has $AB\parallel DC$ and $AD\parallel BC$ (arrows on sides), so it's a parallelogram, so $AD = BC$, $DC = AB$, $\angle D=\angle B$, $\angle A=\angle C$. So if we consider $\triangle ADC$ and $\triangle CBA$:
        - We have $AC$ as common. $\angle D=\angle B$ (opposite angles of parallelogram). $\angle DAC=\angle BCA$ (alternate interior angles, $AD\parallel BC$, $AC$ transversal). Then $AD = BC$ (opposite sides of parallelogram). So with $\angle D=\angle B$, $\angle DAC=\angle BCA$, and $AD = BC$, we can apply AAS. Similarly, with $\angle D=\angle B$, $\angle DCA=\angle BAC$, and $DC = AB$, we can apply AAS.

- So the information needed is: One pair of equal angles (e.g., $\angle D=\angle B$), one more pair of equal angles (e.g., $\angle DAC=\angle BCA$ or $\angle DCA=\angle BAC$) and a pair of equal non - included sides (e.g., $AD = BC$ or $DC = AB$). But since the figure is a parallelogram (from the parallel arrows), we know that $AD = BC$, $DC = AB$, $\angle D=\angle B$. So the key is that we need two angles and a non - included side. Since $AC$ is common, and we have $\angle D=\angle B$ (from parallelogram), we need either $\angle DAC=\angle BCA$ and $AD = BC$ or $\angle DCA=\angle BAC$ and $DC = AB$. But more precisely, in the triangles $\triangle ACD$ and $\triangle CAB$, to use AAS, we need:
        - Two angles (e.g., $\angle D$ and $\angle DAC$ in $\triangle ACD$ and $\angle B$ and $\angle BCA$ in $\triangle CAB$) and the non - included sides ($AD$ and $BC$) to be equal. Or two angles ($\angle D$ and $\angle DCA$ in $\triangle ACD$ and $\angle B$ and $\angle BAC$ in $\triangle CAB$) and the non - included sides ($DC$ and $AB$) to be equal.

- Since $AB\parallel DC$ and $AD\parallel BC$ (arrows), so it's a parallelogram, so $AD = BC$, $DC = AB$, $\angle D=\angle B$. So if we take $\angle D=\angle B$, $\angle DAC=\angle BCA$, and $AD = BC$ (which is true for parallelogram), we can use AAS. So the information needed is that $\angle D=\angle B$, $\angle DAC=\angle BCA$ (or $\angle DCA=\angle BAC$) and $AD = BC$ (or $DC = AB$). But since $AD = BC$ and $DC = AB$ are properties of parallelogram (from parallel sides), the main additional information (if we didn't know it's a parallelogram) would be that $\angle D=\angle B$, one of the alternate interior angles (like $\angle DAC=\angle BCA$) and $AD = BC$. But since the figure shows parallel sides (arrows), we can infer it's a parallelogram, so we know $AD = BC$, $DC = AB$, $\angle D=\angle B$. So to prove congruence by AAS, we need two angles and a non - included side. Since $AC$ is common, and we have $\angle D=\angle B$ (from parallelogram), we need one more angle (e.g., $\angle DAC=\angle BCA$) and the non - included side ( $AD = BC$ which is already true for parallelogram). So the key is that we need to know that two angles and a non - included side are congruent. Given the diagram, since $AB\parallel DC$ and $AD\parallel BC$ (arrows), we know $\angle D=\angle B$, $AD = BC$, and we can get $\angle DAC=\angle BCA$ (alternate interior angles). So with $\angle D=\angle B$, $\angle DAC=\angle BCA$, and $AD = BC$, we can apply AAS. So the information needed is that $\angle D=\angle B$, $\angle DAC=\angle BCA$ (or $\angle DCA=\angle BAC$) and $AD = BC$ (or $DC = AB$). But more simply, since it's a parallelogram, we know $AD = BC$, $DC = AB$, $\angle D=\angle B$. So to use AAS, we need two angles and a non - included side. So we need $\angle D=\angle B$, $\angle DAC=\angle BCA$ (or $\angle DCA=\angle BAC$) and $AD = BC$ (or $DC = AB$). But the most direct answer is that we need to know that $\angle D=\angle B$, $\angle DAC=\angle BCA$ (or $\angle DCA=\angle BAC$) and $AD = BC$ (or $DC = AB$). Since $AD = BC$ and $DC = AB$ are opposite sides of the parallelogram (from parallel sides), and $\angle D=\angle B$ is an opposite angle, and $\angle DAC=\angle BCA$ (or $\angle DCA=\angle BAC$) are alternate interior angles, we can use these to apply AAS.

## Step2: Summarize the AAS Requirement
AAS requires two angles and a non - included side to be congruent between the two triangles. In the given parallelogram - like figure (with $AB\parallel DC$ and $AD\parallel BC$ from the arrows), we know that $AC$ is common, $\angle D=\angle B$ (opposite angles of parallelogram), and we need either:
- $\angle DAC=\angle BCA$ (alternate interior angles) and $AD = BC$ (opposite sides of parallelogram), or
- $\angle DCA=\angle BAC$ (alternate interior angles) and $DC = AB$ (opposite sides of parallelogram)

So the information needed is two angles (e.g., $\angle D$ and $\angle DAC$ in $\triangle ACD$ and $\angle B$ and $\angle BCA$ in $\triangle CAB$) and the non - included sides ( $AD$ and $BC$) to be congruent (or the other pair of angles and non - included sides). Since $AD = BC$ and $DC = AB$ (opposite sides of parallelogram) and $\angle D=\angle B$ (opposite angles of parallelogram), and $\angle DAC=\angle BCA$ (or $\angle DCA=\angle BAC$) (alternate interior angles), we can use these to satisfy AAS.

# Answer:
To prove $\triangle ACD\cong\triangle CAB$ (the two triangles formed by diagonal $AC$) using AAS, we need:
- Two pairs of congruent angles (e.g., $\boldsymbol{\angle D=\angle B}$ (opposite angles of the parallelogram) and $\boldsymbol{\angle DAC=\angle BCA}$ (alternate interior angles, since $AD\parallel BC$) or $\boldsymbol{\angle DCA=\angle BAC}$ (alternate interior angles, since $AB\parallel DC$))
- One pair of congruent non - included sides (e.g., $\boldsymbol{AD = BC}$ or $\boldsymbol{DC = AB}$ (opposite sides of the parallelogram)).

(In the context of the diagram, since $AB\parallel DC$ and $AD\parallel BC$ (arrows indicate parallel sides, so it is a parallelogram), $AD = BC$, $DC = AB$, and $\angle D=\angle B$ are already implied. Thus, confirming $\angle DAC=\angle BCA$ (or $\angle DCA=\angle BAC$) along with the existing equal angles and sides satisfies AAS.)

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Explanation:

Step1: Recall AAS Congruence

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