A circle can be circumscribed about a quadrilateral if and only if both pairs of its opposite angles are supplementary.

An isosceles trapezoid has this characteristic. Assume without loss of generality that  and  are the pairs of base angles. 

Then, since base angles are congruent,  and . Since the bases of a trapezoid are parallel, from the Same-Side Interior Angles Theorem,  and  are supplementary, and, subsequently, so are   and , as well as  and  .

If , then  and  form a supplementary pair, as their measures total ; since the measures of the angles of a quadrilateral total , the measures of   and  also total , making them supplementary as well.

Therefore, it follows from either statement that both pairs of opposite angles are supplementary, and that a circle can be circumscribed about the quadrilateral.

To prove that Parallelogram  is also a rectangle, we need to prove that any one of its angles is a right angle.

If we assume Statement 1 alone, that , then, since  and  form a linear pair,  is right. 

If we assume Statement 2 alone, that , it follows from the converse of the Pythagorean Theorem that  is a right triangle with right angle .

Either way, we have proved that the parallelogram is a rectangle.

It is necessary and sufficient to prove that one of the angles of the parallelogram is a right angle.

Assume Statement 1 alone - that .  and  are supplementary, since they are same-side interior angles of parallel lines. Since ,  is also supplementary to . But as corresponding angles of parallel lines, . Two angles that are conruent and supplementary are both right angles, so  is a right angle.

Assume Statement 2 alone -  that  and  are complementary angles, or, equivalently, . Since the angles of a triangle have measures that add up to , the third angle of , which is , measures , and is a right angle.

Either statement alone proves  a right angle and subsequently proves  a rectangle.

Assume Statement 1 alone. By congruence,  and , making Quadrilateral  a parallelogram. However, no clue is given to whether any angles are right or not, so whether the quadrilateral is a rectangle or not remains open.

Assume Statement 2 alone. By congruence, opposite sides , but no clue is provided as to the lengths of opposite sides  and  . Also, , but no clue is provided as to whether the angles are right. A rectangle would have both characteristics, but so would an isosceles trapezoid with legs  and .

Assume both statements are true. Quadrilateral  is a parallelogram as a consequence of Statement 1. Since   and  are consecutive angles of the parallelogram, they are supplementary, but they are also congruent as a consequence of Statement 2. Therefore, they are right angles, and a parallelogram with right angles is a rectangle.

Statement 1 alone is insufficient to answer the question; A quadrilateral in which   and  are right angles, , and  fits the statement, as well as a rectangle, which by defintion has four right angles. 

Statement 2 alone is insufficient as well, as a parallelogram with acute and obtuse angles, as well as a rectangle, fits the description.

Assume both statements, and construct diagonal  to form two triangles  and . By Statement 1, both triangles are right with congruent legs , and congruent hypotenuses, both being the same segment . By the Hypotenuse Leg Theorem, . By congruence, . The quadrilateral, having two sets of congruent opposite sides, is a parallelogram; a parallelogram with right angles is a rectangle.

Assume Statement 1 alone. A rhombus being a parallelogram, its opposite angles and its adjacent angles are supplementary. From this fact and Statement 1 alone, it follows that

, , , and .

By definition of a rhombus, all of its sides are congruent. By substitution,

 .

All side proportions hold as well as all angle congruences, so the similarity statement holds.

Assume Statement 2 alone. Construct the diagonals of the rhombuses, as follows:

In each rhombus, the diagonals are each other's perpendicular bisector. If 

then 

Since , both angles being right, it follows via the Side-Angle-Side Smiilarity Theorem that 

,

and , by similarity,

.

By a similar argument, 

,

and by angle addition,

.

As with Statement 1 alone, congruence of one set of corresponding angles in two rhombuses leads to the similarity of the two.

Each statement gives the length of one diagonal of the rhombus. Knowing one diagonal is not enough to give the perimeter of the rhombus.

Knowing the lengths of both diagonals, which is the case if both statements are assumed, is enough to determine the perimeter. The rhombus in question, along with its diagonals, is as shown below:

As marked in the diagram, the diagonals are perpendicular, and they are also are each other's bisector. It follows from the given lengths that  and , so  can be calculated using the Pythagorean Theorem. The perimeter is four times , since all sides of a rhombus are congruent.

To prove two figures similar, we must prove that their corresponding angles are congruent, and that their corresponding sides are in proportion.

Assume both statements are true. We show that they provide insufficient information by examining two scenarios.

If Trapezoid  Trapezoid , then  and , so the conditions of both statements are met; also, since the trapezoids are congruent, they are also similar.

Now examine isosceles trapezoid  below, in which , , , and  are positioned on the bases so that  and .

Since  and , Quadrilateral  is a parallelogram, and ; similarly, . Therefore, Trapezoid  is also isosceles, and the conditions of both statements are met. However, corresponding sides are not in proportion, since , but ; consequently, the trapezoids are not similar.

To prove two figures similar, we must prove that their corresponding angles are congruent, and that their corresponding sides are in proportion.

All four sides of a rhombus are congruent, so it easily follows that corresponding sides of two rhombuses are in proportion, regardless of whether they are similar or not; it is therefore necessary and sufficient to prove that corresponding angles are congruent. Also, since a rhombus is a parallelogram, opposite angles are congruent and consecutive angles are supplementary—that is, their angle measures total . Therefore, it is necessary and sufficient to prove just one pair of corresponding angles congruent.

The two statements together give no information about the measures of any of the angles of the rhombus. Therefore, together, they do not answer the question of whether they are similar or not.

To prove two figures similar, we must prove that their corresponding angles are congruent, and that their corresponding sides are in proportion.

All four sides of a rhombus are congruent, so it easily follows that corresponding sides of two rhombuses are in proportion, regardless of whether they are similar or not; it is therefore necessary and sufficient to prove that coresponding angles are congruent. Also, since a rhombus is a parallelogram, opposite angles are congruent and consecutive angles are supplementary - that is, their angle measures total . Therefore, it is neccessary and sufficient to prove just one pair of corresponding angles congruent.

Assume Statement 1 alone.  is a  angle, so any angle consecutive to it, which includes , is supplementary to it—that is, the angle measures total . This makes  a  angle. Its corresponding angle in Rhombus  is , which is a  angle. Since  and  are noncongruent, it follows that Rhombus  Rhombus . (Note that it can be demonstrated that the rhombuses are similar, but the correct statement is Rhombus  Rhombus .)

Statement 2 alone provides no useful information; the relationship between the areas of the rhombuses is irrelevant.