A circle can be circumscribed about a quadrilateral if and only if both pairs of opposite angles are supplementary. This is not proved or disproved by Statement 1 alone:

Case 1:

This is not a rectangle, and opposite angles are supplementary, so a circle can be constructed to circumscribe the quadrilateral.

Case 2:   

This is not a rectangle, and opposite angles are not supplementary, so a circle cannot be constructed to circumscribe the quadrilateral. 

From Statement 2, however, it follows that a two opposite angles are not a supplementary pair, so a circle cannot be circumscribed about it.

For the diagonals of a quadrilateral to be perpendicular, the quadrilateral must be a kite or a rhombus - in either case, there must be two pairs of adjacent congruent sides. Neither statement alone proves this, but both statements together do.

 and , the diagonals of Parallelogram  , are perpendicular if and only if Parallelogram  is also a rhombus.

Opposite sides of a parallelogram are congruent, so if Statement 1 is assumed, . Parallelogram  a rhombus; subsequently,  . 

The angle measures are irrelevant, so Statement 2 is unhelpful.

From Statement 1 alone, we can calculate ,  since two opposite angles of a quadrilateral inscribed inside a circle are supplementary:

From Statement 2 alone, we can calculate , since the degree measure of an inscribed angle of a circle is half that of the arc it intersects:

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.