Which of the following explain how aeb could be proven similar to dec using the aa similarity postulate

∠AEB ≅ ∠DEC because vertical angles are congruent; rotate ΔCED 180° around point E, then translate point D to point A to confirm ∠EAB ≅ ∠EDC.

Step-by-step explanation:

Given:

Two triangles AEB and CED with E as the common vertex for both the triangles.

To prove: Δ AEB similar to Δ DEC by AA postulate.

In order to prove two triangles similar by AA postulate, we have to show any two corresponding angles of both the triangles congruent to each other.

Here, as E serves as the common vertex angles formed at the vertex of both the triangles are vertically opposite to each other and hence are congruent to each other.

∴ ∠ AEB ≅ ∠ CED  (Vertically opposite angles are congruent)

Now, we have to show either ∠ EAB ≅ ∠ EDC or ∠ EBA ≅ ∠ ECD.

For that, we have to rotate the triangle CED by 180° which makes the vertices C and D to lie on BE and AE respectively.

Now, we translate point D to point A to check ∠ EAB ≅ ∠ EDC.

If ∠ EAB ≅ ∠ EDC, then the triangles are similar by AA postulate as two corresponding angles are congruent.

Option 2 and Option 4 shows one corresponding angle congruent and one corresponding side proportional. So, this does not prove by AA postulate.

Option 1 is reflection. Reflection of triangle CED about FG will make vertex C on AE and vertex D on BE. So, we can't make ∠ EAB ≅ ∠ EDC as they lie on different lines.

Hence, only option 3 is correct.

point a is the center of the circle that passes through points e, f, and g and the center of the circle that passes through points x, y, and z.