The given information is actually inconclusive.

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar. Therefore, we seek to prove two of the following three angle congruence statements:

\(\displaystyle \angle AVB \cong \angle CVD\)

\(\displaystyle \angle ABV \cong \angle CDV\)

\(\displaystyle \angle BAV \cong \angle DCV\)

\(\displaystyle \angle AVB\) and \(\displaystyle \angle CVD\) are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, \(\displaystyle \angle AVB \cong \angle CVD\). 

\(\displaystyle \angle ABV \cong \angle DCV\), but this is not one of the statements we need to prove. Also, without further information - for example, whether \(\displaystyle m\) and \(\displaystyle n\) are parallel, which is not given to us - we have no way to prove either of the other two necessary statements. 

The correct response is "false".

As we are establishing whether or not \(\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF\), then \(\displaystyle A\), \(\displaystyle B\), and \(\displaystyle C\) correspond respectively to \(\displaystyle D\), \(\displaystyle E\), and \(\displaystyle F\).

\(\displaystyle \bigtriangleup DEF\) is an isosceles triangle, so it must have two congruent angles. \(\displaystyle \angle F\) has measure \(\displaystyle 30^{\circ }\), so either \(\displaystyle \angle D\) has this measure, \(\displaystyle \angle E\) has this measure, or \(\displaystyle \angle D \cong \angle E\). If we examine the second case, it immediately follows that \(\displaystyle \angle B \ncong \angle E\). One condition of the similarity of triangles is that all pairs of corresponding angles be congruent; since there is at least one case that violates this condition, it does not necessarily follow that  \(\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF\). This makes the correct response "false".

As we are establishing whether or not \(\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF\), then \(\displaystyle A\), \(\displaystyle B\), and \(\displaystyle C\) correspond respectively to \(\displaystyle D\), \(\displaystyle E\), and \(\displaystyle F\).

A triangle is equilateral (having three sides of the same length) if and only if it is also equiangular (having three angles of the same measure, each of which is \(\displaystyle 60^{\circ }\) ). It follows that all angles of both triangles measure \(\displaystyle 60^{\circ }\).

Specifically, \(\displaystyle \angle A \cong \angle D\) and \(\displaystyle \angle B \cong \angle E\), making two pairs of corresponding angles congruent. By the Angle-Angle Similarity Postulate, it follows that \(\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF\), making the correct answer "true".

The sum of the measures of the interior angles of a triangle is \(\displaystyle 180^{\circ }\), so

\(\displaystyle m \angle A + m \angle B + m \angle C = 180^{\circ }\)

\(\displaystyle m \angle A + m \angle B + m \angle C - m \angle C = 180^{\circ } - m \angle C\)

\(\displaystyle m \angle A + m \angle B = 180^{\circ } - m \angle C\)

Also,

\(\displaystyle m \angle A = m \angle B\),

so

\(\displaystyle m \angle A + m \angle A = 180^{\circ } - m \angle C\)

\(\displaystyle 2 \cdot m \angle A = 180^{\circ } - m \angle C\)

\(\displaystyle \frac{2 \cdot m \angle A }{2}= \frac{180^{\circ } - m \angle C}{2}\)

\(\displaystyle m \angle A = \frac{180^{\circ } - m \angle C}{2}\)

By similar reasoning, it holds that

\(\displaystyle m \angle D = \frac{180^{\circ } - m \angle F}{2}\)

Since \(\displaystyle m \angle F = m \angle C\), by substitution,

\(\displaystyle m \angle D = \frac{180^{\circ } - m \angle C}{2}\)

Therefore, 

\(\displaystyle m \angle A = m \angle D\),

or \(\displaystyle \angle A \cong \angle D\)

This, along with the statement that \(\displaystyle \angle C \cong \angle F\), sets up the conditions of the Angle-Angle Similarity Postulate - if two angles of one triangle are congruent to the two corresponding angles of another triangle, the two triangles are similar. It follows that 

\(\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF\).

However, congruence cannot be proved, since at least one side congruence is needed to prove this. This is not given in the problem. 

Therefore, statement (a) must hold, but not necessarily statement (b).

A triangle has \(\displaystyle 180\) degrees.  An isoceles triangle has one vertex angle and two congruent base angles.

Let \(\displaystyle x\) = the base angle and \(\displaystyle 2x\ +\ 20\) = vertex angle

So the equation to solve becomes \(\displaystyle x\ +\ x\ +\ 2x\ +\ 20 = 180\)

or

 \(\displaystyle 4x\ +\ 20=180\)

so the base angle is \(\displaystyle 40\) and the vertex angle is \(\displaystyle 100\) and the difference is \(\displaystyle 60\).

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of \(\displaystyle 180\) degrees. 

Thus, the solution is:

\(\displaystyle 32+74=106\)

\(\displaystyle 180-106=74\)

\(\displaystyle Check: 74+74+32=180\)

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of \(\displaystyle 180\) degrees. Since this is an obtuse isosceles triangle, the two missing angles must be acute angles.  

Thus, the solution is:

\(\displaystyle 180-98=82\)

\(\displaystyle 82\div2=41\)

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of \(\displaystyle 180\) degrees.

Thus, the solution is:

\(\displaystyle 28+28=56\)

\(\displaystyle 180-56=124\)

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of \(\displaystyle 180\) degrees. Since this is an obtuse isosceles triangle, the two missing angles must be acute angles. 

The solution is:

\(\displaystyle 180-154=26\)

However, \(\displaystyle 26\) degrees is the measurement of both of the acute angles combined.

Each individual angle is \(\displaystyle 26\div2=13\).

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of \(\displaystyle 180\) degrees. Since this is an acute isosceles triangle, all of the interior angles must be acute angles.

The solution is:

\(\displaystyle 53+53=106\)

\(\displaystyle 180-106=74\)