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

$\displaystyle \angle A \cong \angle E$, $\displaystyle \angle B \cong \angle F$, $\displaystyle \angle C\cong \angle G$, and $\displaystyle \angle D \cong \angle H$.

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

$\displaystyle \frac{AB}{EF} = \frac{BC}{FG}= \frac{CD}{GH} = \frac{DA}{HE}$ .

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 

$\displaystyle \frac{AC}{EG} = \frac{BD}{FH}$

then 

$\displaystyle \frac{AC \div 2}{EG \div 2} = \frac{BD \div 2}{FH \div 2}$

$\displaystyle \frac{AM}{EN} = \frac{BM}{FN}$

Since $\displaystyle \angle AMB \cong \angle ENF$, both angles being right, it follows via the Side-Angle-Side Smiilarity Theorem that 

$\displaystyle \bigtriangleup MAB \sim \bigtriangleup NEF$,

and , by similarity,

$\displaystyle m \angle MAB = m \angle NEF$.

By a similar argument, 

$\displaystyle m \angle MAD = m \angle NEH$,

and by angle addition,

$\displaystyle m \angle BAD = m \angle FEH$.

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 $\displaystyle AM = 3$ and $\displaystyle BM = 4$, so $\displaystyle AB$ can be calculated using the Pythagorean Theorem. The perimeter is four times $\displaystyle AB$, 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 $\displaystyle TRAP \cong$ Trapezoid $\displaystyle USBQ$, then $\displaystyle \overline{TP} \cong \overline{UQ}$ and $\displaystyle \overline{RA} \cong \overline{SB}$, so the conditions of both statements are met; also, since the trapezoids are congruent, they are also similar.

Now examine isosceles trapezoid $\displaystyle TRAP$ below, in which $\displaystyle U$, $\displaystyle Q$, $\displaystyle S$, and $\displaystyle B$ are positioned on the bases so that $\displaystyle \overline{TP} || \overline{UQ}$ and $\displaystyle \overline{RA} || \overline{SB}$.

Since $\displaystyle \overline{TP} || \overline{UQ}$ and $\displaystyle \overline{TU} || \overline{PQ}$, Quadrilateral $\displaystyle TUPQ$ is a parallelogram, and $\displaystyle \overline{TP} \cong \overline{UQ}$; similarly, $\displaystyle \overline{RA} \cong \overline{SB}$. Therefore, Trapezoid $\displaystyle USBQ$ is also isosceles, and the conditions of both statements are met. However, corresponding sides are not in proportion, since $\displaystyle \overline{TP} \cong \overline{UQ}$, but $\displaystyle \overline{TR} \ncong \overline{US}$; 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 $\displaystyle 180^{\circ }$. 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 $\displaystyle 180^{\circ }$. Therefore, it is neccessary and sufficient to prove just one pair of corresponding angles congruent.

Assume Statement 1 alone. $\displaystyle \angle R$ is a $\displaystyle 60^{\circ }$ angle, so any angle consecutive to it, which includes $\displaystyle \angle H$, is supplementary to it—that is, the angle measures total $\displaystyle 180^{\circ }$. This makes $\displaystyle \angle H$ a $\displaystyle 120^{\circ }$ angle. Its corresponding angle in Rhombus $\displaystyle SIPN$ is $\displaystyle \angle I$, which is a $\displaystyle 60^{\circ }$ angle. Since $\displaystyle \angle H$ and $\displaystyle \angle I$ are noncongruent, it follows that Rhombus $\displaystyle RHOM \nsim$ Rhombus $\displaystyle SIPN$. (Note that it can be demonstrated that the rhombuses are similar, but the correct statement is Rhombus $\displaystyle RHOM \nsim$ Rhombus $\displaystyle IPNS$.)

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

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 $\displaystyle 180^{\circ }$. Therefore, it is neccessary and sufficient to prove just one pair of corresponding angles congruent.

Assume Statement 1 alone. $\displaystyle \angle R$ and $\displaystyle \angle I$ are not corresponding angles, so their congruence does not prove the rhombuses are similar; however it does not prove they are not similar either; for example, two squares are both rhombuses with four right angles, so they are similar and fit this condition.

Assume Statement 2 alone. $\displaystyle \angle R$ and $\displaystyle \angle M$ are two angles of the same rhombus, so similarity cannot be proved or disproved without information about the other rhombus.

Now assume both statements.  $\displaystyle \angle R$ and $\displaystyle \angle M$ are consecutive angles of Rhombus $\displaystyle RHOM$; a rhombus being a parallelogram, the degree measures of the angles total $\displaystyle 180^{\circ }$. From Statement 2, they are congruent, so each measures $\displaystyle 90 ^{\circ }$. Since $\displaystyle \angle R \cong \angle I$ from Statement 1, $\displaystyle \angle I$ also measures $\displaystyle 90 ^{\circ }$. Since each parallelogram has at least one right angle, each has four right angles. Therefore, corresponding angles are congruent, so the rhombuses are similar.

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

We show that Statement 1 alone provides insufficient information by examining two cases.

Case 1: Trapezoid $\displaystyle TRAP \cong$ Trapezoid $\displaystyle USBQ$, with $\displaystyle \angle T$ a $\displaystyle 120^{\circ}$ angle.

A pair of base angles of an isosceles trapezoid are congruent, so, since $\displaystyle \angle T$ measures $\displaystyle 120^{\circ}$, so does $\displaystyle \angle R$. The leg angles of a trapezoid have measures whose sum is $\displaystyle 1 80^{\circ }$, so $\displaystyle \angle A$ and $\displaystyle \angle P$ measure $\displaystyle 60^{\circ}$. By congruence of corresponding angles, $\displaystyle \angle S \cong \angle R$, so $\displaystyle \angle S$ measures $\displaystyle 120^{\circ}$, and the conditions of Statement 1 are met. Furthermore, two congruent triangles are also similar, so Trapezoid $\displaystyle TRAP \sim$ Trapezoid $\displaystyle USBQ$.

Case 2: Examine the figure below, in which $\displaystyle TR = US$. Note that these congruent upper bases have been superimposed upon each other:

The trapezoids are isosceles since their base angles are congruent; also, the conditions of Statement 1 are met. However, $\displaystyle TR = US$, but $\displaystyle BQ \ne AP$. The sides are not in proportion, so the trapezoids are not similar.

Assume Statement 2 alone. Two congruent trapezoids whose side lengths fit the conditions of the statement are also similar. But examine this diagram, in which $\displaystyle TR = US$, $\displaystyle PT=TR= RA$, and $\displaystyle QU=US= SB$. Note that the congruent upper bases have been superimposed upon each other:

The conditions of Statement 2 are met, but $\displaystyle \angle T \ncong \angle U$, so the trapezoids are not similar.

Now assume both statements to be true. As stated before, it follows from Statement 1 that all corresponding angles are congruent. $\displaystyle PT=TR= RA$ and $\displaystyle QU=US= SB$, so it follows from the Division Property of Equality that

$\displaystyle \frac{PT}{QU}=\frac{TR}{US}= \frac{RA}{SB}$

It remains to be demonstrated that $\displaystyle \frac{AP}{BQ}$ is equal to the above ratios as well. If the diagonals $\displaystyle \overline{RP }$ and $\displaystyle \overline{SQ}$ are constructed, $\displaystyle \bigtriangleup TRP$ and $\displaystyle \bigtriangleup USQ$ are formed; since $\displaystyle \frac{PT}{QU}=\frac{TR}{US}$ and $\displaystyle \angle T \cong \angle U$, by the Side-Angle-Side Similarity Theorem, $\displaystyle \bigtriangleup TRP \sim \bigtriangleup USQ$. It follows that $\displaystyle \angle TRP \cong \angle USQ$, and by angle addition, since $\displaystyle \angle TRA \cong \angle USB$, it follows that $\displaystyle \angle PRA \cong \angle QSB$. From the Angle-Angle Postulate, the other two triangles formed are similar—that is, $\displaystyle \bigtriangleup RAP \sim \bigtriangleup SBQ$—and it follows that $\displaystyle \frac{AP}{BQ}= \frac{RA}{SB}$. Therefore, all four sides of the trapezoids are in proportion, and the trapezoids are similar.

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.

Case 1: Trapezoid $\displaystyle TRAP \cong$ Trapezoid $\displaystyle USBQ$, then $\displaystyle \angle T \cong \angle U$ and $\displaystyle \angle A \cong \angle B$, so the conditions of both statements are met; also, since the trapezoids are congruent, they are also similar.

Now examine isosceles trapezoid $\displaystyle TRAP$ below, in which $\displaystyle U$, $\displaystyle Q$, $\displaystyle S$, and $\displaystyle B$ are positioned on the bases so that $\displaystyle \overline{TP} || \overline{UQ}$ and $\displaystyle \overline{RA} || \overline{SB}$.

Since $\displaystyle \overline{TP} || \overline{UQ}$ and $\displaystyle \overline{TU} || \overline{PQ}$, Quadrilateral $\displaystyle TUPQ$ is a parallelogram, and $\displaystyle \overline{TP} \cong \overline{UQ}$; similarly, $\displaystyle \overline{RA} \cong \overline{SB}$. Therefore, Trapezoid $\displaystyle USBQ$ is also isosceles. Also, by the Corresponding Angles Theorem, $\displaystyle \angle T \cong \angle SUQ$ and $\displaystyle \angle A \cong \angle SBQ$, and the conditions of both statements are met. However, corresponding sides are not in proportion, since $\displaystyle \overline{TP} \cong \overline{UQ}$, but $\displaystyle \overline{TR} \ncong \overline{US}$; consequently, the trapezoids are not similar.

Similar rectangles (or any shape for that matter) are the same shape but can be different sizes. What that means is that their side lengths all follow a common ratio; if one pair of corresponding sides follow the ratio 2:1, then all corresponding sides must follow the same ratio.

Statement I gives us the perimeter and one side of $\displaystyle BPSD$.

Statement II gives us the area and one side of $\displaystyle KMDS$.

We can find all the sides of both rectangles, but we need both statements to do so. Once we have all side lengths, we can compare them to see if they follow the same ratios.

IF $\displaystyle BPSD$ has perimeter of 16 and one side is 3, we can find the other side using the following:

$\displaystyle 16=2l+2\cdot 3$

$\displaystyle l=5$

If $\displaystyle KMDS$ has area of 44 and side of 6, the other side can be found via the following:

$\displaystyle 44=6\cdot w$

$\displaystyle w=\frac{44}{6}=\frac{22}{3}$

Compare the ratios of the sides to find out whether the two rectangles are similar:

$\displaystyle \frac{3}{6}\not\equiv \frac{5}{\frac{22}3}$

Therefore, the rectangles are not similar.