Question

given: \\(\overline{ab} \cong \overline{bc}\\) and \\(\angle abe \cong \angle cbd\\).
prove: \\(\triangle abe \cong \triangle cbd\\).
note: quadrilateral properties are not permitted in this proof.
step statement reason
1 \\(\overline{ab} \cong \overline{bc}\\)
\\(\angle abe \cong \angle cbd\\) given
try type of statement
image of a triangle with vertices a, b, c, and points e, d on ac, with ab ≅ bc marked, and angles abe ≅ cbd marked

Explanation:

To prove the congruence of \( \triangle ABE\) and \( \triangle CBD\), we can use either the SAS or ASA congruence criteria. We start with the given congruent sides (\( \overline{AB}\cong\overline{BC}\)) and angles (\( \angle ABE\cong\angle CBD\)). For SAS, we need to identify another pair of congruent sides (e.g., \( \overline{BE}\cong\overline{BD}\) from the diagram or properties). For ASA, we use the isosceles triangle theorem (since \( \overline{AB}\cong\overline{BC}\), \( \angle A\cong\angle C\)) to get the third part of the ASA criterion. The key is to identify the correct combination of sides and angles to apply a triangle congruence postulate (SAS or ASA) to show \( \triangle ABE\cong\triangle CBD\).

Answer:

To prove \( \triangle ABE \cong \triangle CBD \), we can use the SAS (Side - Angle - Side) congruence criterion. Here are the steps:

Step 1: Identify the given congruent sides and angles

We are given that \( \overline{AB} \cong \overline{BC} \) and \( \angle ABE \cong \angle CBD \).

Step 2: Identify the included side or the other pair of congruent sides

We need to find another pair of congruent sides. Notice that \( \overline{BE} \) and \( \overline{BD} \)? Wait, no. Wait, looking at the diagram, we can also consider the fact that \( \angle A\) and \( \angle C\)? No, wait, let's re - examine.

Wait, actually, if we look at the triangles \( \triangle ABE \) and \( \triangle CBD \):

• Side: \( AB\cong BC\) (given)
• Angle: \( \angle ABE\cong\angle CBD\) (given)
• Side: We need to find \( BE\cong BD\)? Wait, no, maybe \( AE\cong CD\)? Wait, no, the diagram shows that \( AB = BC\) (marked with one tick), and maybe \( BE = BD\)? Wait, no, perhaps the triangles are isosceles. Wait, since \( AB = BC\), \( \triangle ABC\) is isosceles with \( \angle A=\angle C\). But we are not allowed to use quadrilateral properties.

Wait, another approach: Let's check the sides around the congruent angles.

In \( \triangle ABE \) and \( \triangle CBD \):

• \( AB = BC\) (given)
• \( \angle ABE=\angle CBD\) (given)
• We need to show that \( BE = BD\) or \( AE = CD\). Wait, maybe the diagram has \( BE\) and \( BD\) as equal? Wait, no, the problem is to fill in the statement. Wait, the first step is given, then for the next step, we can say that \( \overline{BE}\cong\overline{BD}\)? No, maybe not. Wait, perhaps the triangles share the angle - side relationship.

Wait, actually, the correct pair of sides: Since \( AB = BC\), \( \angle ABE=\angle CBD\), and if we can show that \( BE = BD\), but maybe in the diagram, \( BE\) and \( BD\) are parts of the same line or have equal length. Alternatively, maybe \( \angle A=\angle C\) because \( AB = BC\) (isosceles triangle property: in \( \triangle ABC\), \( AB = BC\), so \( \angle A=\angle C\)). But the note says quadrilateral properties are not permitted. Wait, \( \triangle ABC\) is a triangle, not a quadrilateral. So in \( \triangle ABC\), \( AB = BC\), so \( \angle A=\angle C\) (isosceles triangle theorem: if two sides of a triangle are congruent, then the angles opposite those sides are congruent).

So let's re - structure:

Step 1:

Statement: \( \overline{AB}\cong\overline{BC}\), \( \angle ABE\cong\angle CBD\)
Reason: Given

Step 2:

Statement: \( \angle A\cong\angle C\)
Reason: Isosceles Triangle Theorem (since \( AB = BC\), \( \triangle ABC\) is isosceles with \( \angle A\) and \( \angle C\) as base angles)

Step 3:

Now, in \( \triangle ABE\) and \( \triangle CBD\):

• \( AB = BC\) (given)
• \( \angle A=\angle C\) (proven in step 2)
• \( \angle ABE=\angle CBD\) (given)

Wait, no, that would be ASA (Angle - Side - Angle).

Wait, let's go back. The problem is to fill in the "Type of Statement" for the next step. After the given, we can state that \( \overline{BE}\cong\overline{BD}\) (if that's the case from the diagram) or \( \angle A\cong\angle C\).

But if we use SAS:

• Side: \( AB = BC\)
• Angle: \( \angle ABE=\angle CBD\)
• Side: \( BE = BD\)

If we assume that \( BE = BD\) (maybe from the diagram's markings, although not shown in the given), then by SAS, \( \triangle ABE\cong\triangle CBD\).

Alternatively, if we use ASA:

• Angle: \( \angle A=\angle C\) (from \( AB = BC\), isosceles triangle)
• Side: \( AB = BC\)
• Angle: \( \angle ABE=\angle CBD\)

So, to complete the proof, after the given, we can add the statement that \( \angle A\cong\angle C\) (by isosceles triangle theorem) or \( BE\cong BD\) (if applicable from the diagram).

If we follow the SAS path:

1. \( \overline{AB}\cong\overline{BC}\) (given)
2. \( \angle ABE\cong\angle CBD\) (given)
3. \( \overline{BE}\cong\overline{BD}\) (assumed or from diagram)
4. \( \triangle ABE\cong\triangle CBD\) (SAS)

If we follow the ASA path:

1. \( \overline{AB}\cong\overline{BC}\) (given)
2. \( \angle ABE\cong\angle CBD\) (given)
3. \( \angle A\cong\angle C\) (isosceles triangle theorem, since \( AB = BC\))
4. \( \triangle ABE\cong\triangle CBD\) (ASA)

Since the problem is to fill in the "Type of Statement" for the next step (after step 1), the most appropriate statement would be either \( \angle A\cong\angle C\) (using isosceles triangle theorem) or \( \overline{BE}\cong\overline{BD}\) (if applicable).

Assuming we use the isosceles triangle theorem, the next statement is \( \angle A\cong\angle C\) with the reason "Isosceles Triangle Theorem (If two sides of a triangle are congruent, then the angles opposite those sides are congruent)".