If $\displaystyle m\angle A =m \angle B = 60^{\circ }$, then 

$\displaystyle m\angle C = 180 - \left (m\angle A + m \angle B \right )=180 - \left (60 +60 \right ) = 60^{\circ }$.

This makes $\displaystyle \Delta ABC$ an equiangular triangle.

If $\displaystyle \Delta ABC \sim \Delta DEF$, and $\displaystyle \Delta DEF$ is equiangular, then, since corresponding angles of similar triangles are congruent, $\displaystyle \Delta ABC$ has the same angle measures, and is itself equiangular.

From either statement, since all equiangular triangles are equilateral, we can draw this conclusion about $\displaystyle \Delta ABC$.

The two statements together provide insufficient information. A triangle with sides $\displaystyle 4$, $\displaystyle 4$, and $\displaystyle 4$ is equilateral and has perimeter $\displaystyle 12$; A triangle with sides $\displaystyle 3$, $\displaystyle 4$, and $\displaystyle 5$ is not equilateral and has perimeter $\displaystyle 12$.

To find the answer we should know more about the characteristics of the triangle, i.e. its angles, sides...

Statement 1 alone is obviously insufficient, since we don't know whether the triangle is equilateral, nothing can be said about AB.

Statement 2 is equally as unhelpful as statement 1, since we don't know whether ABC is of a specific type of triangle.

Taken together, these statements allow us to calculate the length of CB, but we can't go further, because we don't know what is AD.

Therefore statements 1 and 2 are not sufficient even taken together.

To find the length of the side, we would need to know anything about the lengths in the circle or in the triangle.

From statement 1, we can find the radius of the circle, which allows us to calculate the height of the triangle, since the radius is $\displaystyle \frac{2}{3}$ of the height. And finally since the triangle is equilateral, we can also calculate the length of the sides from the height.

Therefore statement 1 is sufficient.

Statement 2 also gives us useful information, indeed the perimeter is simply three times the length of the sides.

Therefore the final answer is each statement alone is sufficient.

I) Tells us the perimeter of the triangle. 

II) Tells us that FHT is an equilateral triangle.

Taking these statements together we are able to find the side length by dividing the perimeter from statement I, by 3 since all side lengths of an equilateral are the same by statement II.

$\displaystyle s=\frac{P}{3} \rightarrow s=\frac{114}{3}=38ft$

Assume Statement 1 alone. Since all three sides of $\displaystyle \bigtriangleup ABC$ are congruent - specifically, $\displaystyle \overline{AB} \cong \overline{BC}$ - and $\displaystyle \overline{BC} \cong \overline{EF}$, it follows by transitivity that $\displaystyle \overline{AB} \cong \overline{EF}$. However, no information is given as to whether $\displaystyle \overline{DE}$ has length greater then, equal to, or less than $\displaystyle \overline{EF}$, so which of $\displaystyle \overline{AB}$ and $\displaystyle \overline{DE}$, if either, is the longer cannot be answered.

Assume Statement 2 alone. Since $\displaystyle \angle D$ is the right angle of $\displaystyle \bigtriangleup DEF$, $\displaystyle \overline{EF}$ is the hypotenuse and this the longest side, so $\displaystyle EF > DE$ and $\displaystyle EF > DF$. However, no comparisons with the sides of $\displaystyle \bigtriangleup ABC$ can be made.

Now assume both statements are true. $\displaystyle AB = EF$ as a consequence of Statement 1, and $\displaystyle EF > DE$ as a consequence of Statement 2, so $\displaystyle AB > DE$.

An equilateral triangle has three sides of equal measure, so if the length of any one of the three sides can be determined, the lengths of all three can be as well.

Assume Statement 1 alone. $\displaystyle \overline{BC}$ is a diagonal of a rectangle of area 30. However, neither the length nor the width can be determined, so the length of this segment cannot be determined with certainty.

Assume Statement 2 alone. A square with area 36 has sidelength the square root of this, or 6; its diagonal, which is $\displaystyle \overline{AC}$, has length $\displaystyle \sqrt{2}$ times this, or $\displaystyle 6 \sqrt{2}$. This is also the length of $\displaystyle \overline{AB}$.

We demonstrate that either statement alone yields sufficient information by noting that the circle that includes all three vertices of a triangle - described in Statement 1 - is its circumscribed circle, and that the circle that includes all three midpoints of the sides of an equilateral triangle - described in Statement 2 - is its inscribed circle. We examine this figure below, which shows the triangle, both circles, and the three altitudes:

The three altitudes intersect at $\displaystyle O$, which divides each altitude into two segments whose lengths have ratio 2:1. $\displaystyle O$ is the center of both the circumscribed circle, whose radius is $\displaystyle OC$, and the inscribed circle, whose radius is $\displaystyle OM$.

Therefore, from Statement 1 alone and the area formula for a circle, we can find $\displaystyle OC$ from the area $\displaystyle 576\pi$ of the circumscribed circle:

$\displaystyle \pi (OC)^{2} =a$

$\displaystyle \pi (OC)^{2} = 576 \pi$

$\displaystyle (OC)^{2} =576$

$\displaystyle OC= 24$

From Statement 2 alone and the circumference formula for a cicle, we can find $\displaystyle OM$ from the circumference $\displaystyle 24\pi$ of the inscribed circle:

$\displaystyle 2 \pi \cdot OM = c$

$\displaystyle 2 \pi \cdot OM = 24 \pi$

$\displaystyle OM =12$

By symmetry, $\displaystyle \bigtriangleup COM$ is a 30-60-90 triangle, and either way, $\displaystyle CM = 12\sqrt{3}$, and $\displaystyle BC =2 \cdot CM =2 \cdot 12\sqrt{3}= 24 \sqrt{3}$.

An equilateral triangle has three sides of equal length, so $\displaystyle AB = AC = BC$ and $\displaystyle DE = EF = DF$.

Assume Statement 1 alone. Since $\displaystyle AC> EF$, then, by substitution, $\displaystyle AB > DE$.

Assume Statement 2 alone. Since $\displaystyle BC = DF + 1$, it follows that $\displaystyle BC > DF$, and again by substitution, $\displaystyle AB > DE$.

Assume Statement 1 alone, and let $\displaystyle P$ be the common perimeter of the triangles. Since an equilateral triangle has three sides of equal length, $\displaystyle AB = \frac{1}{3} P$ and $\displaystyle DE = \frac{1}{3} P$, so $\displaystyle AB = DE$.

Assume Statement 2 alone, and let $\displaystyle A$ be the common area of the triangles. Using the area formula for an equilateral triangle, we can note that:

$\displaystyle A = \frac{s^{2}\sqrt{3}}{4}$

$\displaystyle A = \frac{ (AB)^{2}\sqrt{3}}{4}$ and $\displaystyle A = \frac{ (DE)^{2}\sqrt{3}}{4}$,  

so

$\displaystyle \frac{ (AB)^{2}\sqrt{3}}{4}= \frac{ (DE)^{2}\sqrt{3}}{4}$

$\displaystyle \frac{ (AB)^{2}\sqrt{3}}{4}\cdot \frac{4}{\sqrt{3}}= \frac{ (DE)^{2}\sqrt{3}}{4} \cdot \frac{4}{\sqrt{3}}$

$\displaystyle (AB)^{2} = (DE)^{2}$

$\displaystyle AB = DE$.