For simplicity's sake, we will assume that  has sidelength 1, and, consequently, perimeter 3; these arguments work regardless of the size of the triangle.

We will also need the circumference formula .

Assume Statement 1 alone. Since the midpoint of , which we will call , is inside the circle, the radius of the circle must be greater than  . This makes the circumference at least  times this, or , which is greater than 3.

Assume Statment 2 alone. Since the circle has as a radius the segment from  to the midpoint of the opposite side, it is an altitude of , and the radius is the height of the triangle. By way of the 30-60-90 Theorem, this height is , and the circumference of the circle is  times this, or . This is greater than 3.

Either statement alone establishes that the circumference of the circle is greater than 3, the perimeter of .

Assume both statements to be true. From Statement 1, since , it follows that ; since the perimeter of an equilateral triangle is three times the length of one side, it follows that  has perimeter greater than . Similarly, from Statement 2, it follows that  has perimeter greater than . However, there is no way to determine whether  or  has the greater perimeter of the two.

Assume Statement 1 alone, and examine the diagram below, which shows a circle circumscribed about :

The diameter, which is equal to  as given by Statement 1, is greater in length than any chord which is not a diameter - and all sides of  are non-diameter chords. Therefore,  has sides of greater length than , and its perimeter is therefore greater. However, nothing is given about . 

If Statement 2 alone is assumed, then, similarly,  can be shown to have perimeter greater than that of . But nothing can be determined about .

From the two statements together, however,  has a perimeter greater than those of the other two triangles.

If we let  and  be the sidelengths of  and , respectively; the statements can be rewritten as:

Statement 1: 

Statement 2: 

Since the perimeter of an equilateral triangle is three times its common sidelength, comparison of the lengths of the sides is all that is necessary to determine which triangle, has the greater perimeter. The question can therefore be reduced to asking which of  and , if either, is greater.

Statement 1 alone is not sufficient to yield an answer:

Case 1: 

Case 2: 

Both cases satisfy Statement 1, but in the first case, , meaning that  has greater sidelength and perimeter than , and in the second case, , meaning the reverse. By a similar argument, Statement 2 is insufficient.

Now assume both statements to be true. The two equations together comprise a system of equations:

Multiply the first equation by 3 and the second by , then add:

Now substitute back:

 has the greater sidelength and, consequently, the greater perimeter.

Assume Statement 1 alone. Since in ,  is the right angle,  is the hypotenuse, which is longer than either of the two legs -  and . Since, in equilateral triangle , , it follows that  and . Consequently

,

and  has the greater perimeter.

Statement 2 alone gives insufficient information, since we know nothing about the second leg or the hypotenuse of . We see this by examining these two cases.

Case 1: Let  be an equilateral triangle of sidelength 6, and  in . If second leg  of  has length 4, then the triangle  is a 3-4-5 triangle, with hypotenuse  having length 5. The perimeter of  is , and the perimeter of  is .  has the greater perimeter.

Case 2: Let  be an equilateral triangle of sidelength 6, and  in . If second leg  of  has length 20,  has the greater perimeter on the basis of one side alone.

Recall the formula for perimeter of a triangle.

where  represents the side length of the triangle

Statement 1: We're given the length of the side so all we need to do is plug this value into the equation. 

Statement 2: We're given the area so we first need to solve for the side length.

Now we can plug the value into the equation, like we did in Statement 1.

To prove triangle congruence, we need to establish some conditions involving side and angle congruence.

We know  from reflexivity.

If we only assume Statement 1, then we know that , both angles being right angles. If we only assume Statement 2, then we know that  by definition of a bisector. Either way, we only have one angle congruence and one side congruence, not enough to establish congruence between triangles. The two statements together, however, set up the Angle-Side-Angle condition, which does prove that .

To prove triangle congruence, we need to establish some conditions involving side and angle congruence.

We know  from reflexivity.

From Statement 1 alone, by definition,  and, since both  and  are right angles, . This sets the conditions to apply the Side-Angle-Side Postulate to prove that .

From Statement 2 alone,we know that  by definition of a bisector. But we only have one angle congruence and one side congruence, not enough to establish congruence between triangles. 

Assume Statement 1 alone. If , then  and . That  does not prove or disprove the congruence statement. Therefore, Statement 1 alone - and by a similar argument, Statement 2 alone - is not sufficient to answer the question.

Now assume both statements to be true.

Suppose . Then this, along with the two statements, can be combined to yield the statement

.

Similarly, if ,  

,

and .

 and  cannot both be true, so it is impossible for .

From Statement 1 alone, we are only give one angle congruency and one side congruency, which, without further information, is not enough to prove or to disprove triangle congruence. By a similar argument, Statement 2 alone is not sufficient either.

Assume both statements are true. By Statement 1, the hypotenuses are congruent, and by Statement 2, one pair of corresponding legs are congruent. These are the conditions of the Hypotenuse Leg Theorem, so  is proved to be true.