Figure 6. Given △ABC and △EDB, ∠BAC = ∠BED, AB = ED, and AC = EB. Prove that △ABC ≅ △EDB.

Proof:

We are given two triangles, △ABC and △EDB, with the following information:

• ∠BAC = ∠BED (Given)
• AB = ED (Given)
• AC = EB (Given)

We need to prove that △ABC ≅ △EDB using the congruence postulates (SSS, SAS, ASA, AAS, HL).

Let's analyze the given information in relation to the vertices of the triangles:

• Side AB corresponds to side ED.
• Side AC corresponds to side EB.
• Angle ∠BAC is between sides AB and AC in △ABC.
• Angle ∠BED is between sides ED and EB in △EDB.

We have two sides and the angle included between them in △ABC (AB, AC, ∠BAC) and corresponding parts in △EDB (ED, EB, ∠BED).

Specifically, we have:

1. AB = ED (Given Side)
2. AC = EB (Given Side)
3. ∠BAC = ∠BED (Given Angle)

The angle ∠BAC is included between sides AB and AC in △ABC. The angle ∠BED is included between sides ED and EB in △EDB.

Therefore, by the Side-Angle-Side (SAS) congruence postulate, if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the triangles are congruent.

Conclusion: △ABC ≅ △EDB by SAS congruence.