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It may not be understood at first glance, but all squares, rectangles, and rhombi are parallelograms. Therefore, all the properties of parallelograms apply to these quadrilaterals as well! In this lesson, theorems about parallelograms will be discussed to understand this concept better.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

In the applet below, a parallelogram, a rectangle, a rhombus, and a square can be selected and rotated $90_{∘}$ clockwise about the point of intersection of its diagonals. What can be noted when these polygons are rotated $180_{∘}?$
*parallelograms*. Additionally, when rotated $180_{∘},$ each quadrilateral is mapped onto itself. Considering this information, what conclusions can be made about the opposite sides of a parallelogram?

Note that all the quadrilaterals in the applet are

A conclusion that can be made from the previous exploration is that the opposite sides of a parallelogram are congruent. This is explained in detail in the following theorem.

The opposite sides of a parallelogram are congruent.

In respects to the characteristics of the diagram, the following statement holds true.

$PQ ≅SRandQR ≅PS$

This theorem can also be proven by using congruent triangles. Consider the parallelogram $PQRS$ and its diagonal $PR.$

It can be noted that two triangles are formed with $PR$ as a common side.$△PQRand△RSP $

By the definition of a parallelogram, $PQ $ and $SR$ are parallel. Therefore, by the Alternate Interior Angles Theorem, it can be stated that $∠QPR≅∠SRP$ and that $∠QRP≅∠SPR.$ Furthermore, by the Reflexive Property of Congruence, $PR$ is congruent to itself.
Consequently, $△PQR$ and $△RSP$ have two pairs of congruent angles and an included congruent side. $∠QPRPR∠QRP ≅∠SRP≅PR≅∠SPR $

Therefore, by the Angle-Side-Angle Congruence Theorem, $△PQR$ and $△RSP$ are congruent triangles.
$△PQR≅△RSP $

Since corresponding parts of congruent figures are congruent, $PS$ is congruent to $QR $ and $PQ $ is congruent to $RS.$ $PQ ≅SRandQR ≅PS$

Furthermore, it can be stated whether a quadrilateral is a parallelogram just by checking if its opposite sides are congruent.

If the opposite sides of a quadrilateral are congruent, then the polygon is a parallelogram.

Following the above diagram, the statement below holds true.

If $PQ ≅SR$ and $QR ≅PS,$ then $PQRS$ is a parallelogram.

This theorem can be proven by using congruent triangles. Consider the quadrilateral $PQRS,$ whose opposite sides are congruent, and its diagonal $PR.$ By the Reflexive Property of Congruence, this diagonal is congruent to itself.

Therefore, by the Side-Side-Side Congruence Theorem, $△PQR$ and $△RSP$ are congruent triangles.$⎩⎪⎪⎨⎪⎪⎧ PQ ≅RSQR ≅SPPR≅PR ⇒△PQR≅△RSP $

Since corresponding parts of congruent figures are congruent, corresponding angles of $△PQR$ and $△RSP$ are congruent.
Finally, by the Converse of the Alternate Interior Angles Theorem, $PQ $ is parallel to $RS$ and $QR $ is parallel to $SP.$ Therefore, by the definition of a parallelogram, $PQRS$ is a parallelogram.

This proves the theorem.

If $PQ ≅SR$ and $QR ≅PS,$ then $PQRS$ is a parallelogram.

Another conclusion that can be made from the exploration is that the opposite angles of a parallelogram are congruent. This is explained in detail in the following theorem.
### Rule

## Parallelogram Opposite Angles Theorem

### Proof

### Rule

## Converse Parallelogram Opposite Angles Theorem

### Proof

Since $x+y=180,$ the consecutive interior angles of $ABCD$ are supplementary. Therefore, by the Converse Consecutive Interior Angles Theorem, it can be concluded that opposite sides of $ABCD$ are parallel.
#### Completed Proof

**Proof:**

In a parallelogram, the opposite angles are congruent.

For the parallelogram $PQRS,$ the following statement holds true.

$∠Q≅∠Sand∠P≅∠R$

This theorem can be proved by using congruent triangles. Consider the parallelogram $PQRS$ and its diagonal $PR.$

Opposite sides of a parallelogram are parallel. Therefore, by the Alternate Interior Angles Theorem it can be stated that $∠QPR≅∠SRP$ and $∠QRP≅∠SPR.$ Furthermore, by the Reflexive Property of Congruence, $PR$ is congruent to itself.

Two angles of $△PQR$ and their included side are congruent to two angles of $△RSP$ and their included side. By the Angle-Side-Angle Congruence Theorem, $△PQR$ and $△RSP$ are congruent triangles.$△PQR≅△RSP $

Since corresponding parts of congruent figures are congruent, $∠Q$ and $∠S$ are congruent angles. By drawing the diagonal $QS $ and using a similar procedure, it can be shown that $∠P$ and $∠R$ are also congruent angles.

$∠Q≅∠Sand∠P≅∠R$

Furthermore, it can be determined whether a quadrilateral is a parallelogram just by looking at its opposite angles.

If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

Based on the above diagram, the following statement holds true.

$If$ $∠A≅∠C$ $and$ $∠B≅∠D,$ $then$ $ABCD$ $is$ $a$ $parallelogram.$

Assume that $ABCD$ is a quadrilateral with opposite congruent angles. It should be noted that congruent angles have the same measure. Then, let $x_{∘}$ be the measure of $∠A$ and $∠C,$ and $y_{∘}$ be the measure of $∠B$ and $∠D.$

By the Polygon Interior Angles Theorem, the sum of the interior angles of a quadrilateral is $360_{∘}.$ With this information, a relation can be found between the consecutive interior angles of $ABCD.$$x+y+x+y=360$

$x+y=180$

Consequently, by the definition of a parallelogram, $ABCD$ is a parallelogram.

$ Given:Prove: ∠A≅∠Cand∠B≅∠DABCDis a parallelogram. $

To be able to be carefree and enjoy a soccer match over the weekend, Vincenzo wants to complete his Geometry homework immediately after school. He is given a diagram showing a parallelogram, and asked to find the values of $a,$ $b,$ and $x.$

Find the values of $a,$ $b,$ and $x$ to help Vincenzo be carefree for the match!{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">a<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.674115em;vertical-align:0em;\"><\/span><span class=\"mord\"><span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.674115em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mbin mtight\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["30"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.69444em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">b<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.674115em;vertical-align:0em;\"><\/span><span class=\"mord\"><span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.674115em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mbin mtight\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["10"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["5"]}}

First, for simplicity, the value of $x$ will be found. After that, the values of $a$ and $b$ will be calculated.

$2x+5=5x−10 $

This equation can now be solved for $x.$
$2x+5=5x−10$

$x=5$

${(5a+10)_{∘}=(10b+60)_{∘}(a−10)_{∘}=(3b−10)_{∘} (I)(II) $

This system will be solved by using the Substitution Method. For simplicity, the degree symbol will be removed.
${5a+10=10b+60a−10=3b−10 $

Solve by substitution

AddEqn

$(II):$ $LHS+10=RHS+10$

${5a+10=10b+60a=3b $

Substitute

$(I):$ $a=3b$

${5(3b)+10=10b+60a=3b $

Multiply

$(I):$ Multiply

${15b+10=10b+60a=3b $

SubEqn

$(I):$ $LHS−10=RHS−10$

${15b=10b+50a=3b $

SubEqn

$(I):$ $LHS−10b=RHS−10b$

${5b=50a=3b $

DivEqn

$(I):$ $LHS/5=RHS/5$

${b=10a=3b $

Substitute

$(II):$ $b=10$

${b=10a=3(10) $

Multiply

$(II):$ Multiply

${b=10a=30 $

In the following applet, a parallelogram is shown. By dragging one of its vertices, different types of parallelograms such as squares, rectangles, and rhombi can be formed. By using the measuring tool provided, investigate what relationships exist between the diagonals of each parallelogram.

After investigating each parallelogram type, what relationships between the diagonals of the parallelograms were discovered?

A conclusion that can be made from the previous exploration is that the diagonals of a parallelogram intersect at their midpoint. This is explained in detail in the following theorem.

In a parallelogram, the diagonals bisect each other.

If $PQRS$ is a parallelogram, then the following statement holds true.

$PM≅RMandQM ≅SM$

This theorem can be proven by using congruent triangles. Consider the parallelogram $PQRS$ and its diagonals $PR$ and $QS .$ Let $M$ be the point intersection of the diagonals.

Since $PQ $ and $SR$ are parallel, by the Alternate Interior Angles Theorem it can be stated that $∠QPR≅∠SRP$ and that $∠PQS≅∠RSQ.$ Furthermore, by the Parallelogram Opposite Sides Theorem it can be said that $PQ ≅SR.$

Here, two angles of $△PMQ$ and their included side are congruent to two angles of $△RMS$ and their included side. Therefore, by the Angle-Side-Angle Congruence Theorem $△PMQ$ and $△RMS$ are congruent triangles.$△PMQ≅△RMS $

Since corresponding parts of congruent triangles are congruent, $PM$ is congruent to $RM$ and $QM $ is congruent to $SM.$ $PM≅RMandQM ≅SM$

By the definition of a segment bisector, both segments $PR$ and $QS $ are bisected at point $M.$ Therefore, it has been proven that the diagonals of a parallelogram bisect each other.

Also, a quadrilateral can be identified as a parallelogram just by looking at its diagonals.

If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Based on the diagram above, the following relation holds true.

If $AC$ and $BD$ bisect each other, then $ABCD$ is a parallelogram.

Let $E$ be point of intersection of the diagonals of a quadrilateral. Since the diagonals bisect each other, $E$ is the midpoint of each diagonal.

Because $∠AEB$ and $∠CED$ are vertical angles, they are congruent by the Vertical Angles Theorem. Therefore, by the Side-Angle-Side Congruence Theorem, $△AEB$ and $△CED$ are congruent triangles. Since corresponding parts of congruent figures are congruent, $AB$ and $CD$ are congruent.

Applying a similar reasoning, it can be concluded that $△AED$ and $△CEB$ are congruent triangles. Consequently, $AD$ and $BC$ are also congruent.

Finally, since both pairs of opposite sides of quadrilateral $ABCD$ are congruent, the Converse Parallelogram Opposite Sides Theorem states that $ABCD$ is a parallelogram.

Vincenzo has one last exercise to finish before going to a soccer match. He has been given a diagram showing a parallelogram. He is asked to find the value of $x$ and $y.$

Find the values of $x$ and $y$ and help Vincenzo finish his homework!{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["4"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.625em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">y<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["2"]}}

According to the Parallelogram Diagonals Theorem, the diagonals of a parallelogram bisect each other.

With this information, two equations can be written.$ (I)9=2x+1(II)y+4=3y $

These can be solved one at a time. Equation (I) will be solved first.
$9=2x+1$

Solve for $x$

$x=4$

$y+4=3y$

Solve for $y$

$y=2$

Consider a rigid motion that rotates a parallelogram $180_{∘}$ about the point of intersection of its diagonals.

Since a rotation is a rigid motion, the preimage and the image are congruent figures. Furthermore, because corresponding parts of congruent figures are congruent, the three statements below hold true.

- Opposite sides of a parallelogram are congruent.
- Opposite angles of a parallelogram are congruent.
- The diagonals of a parallelogram bisect each other.

It can be determined whether a parallelogram is a rectangle just by looking at its diagonals. Furthermore, if a parallelogram is a rectangle, a statement about its diagonals can be made.

A parallelogram is a rectangle if and only if its diagonals are congruent.

Based on the diagram, the following relation holds true.

$PQRS$ is a rectangle $⇔$ $PR≅QS $

Two proofs will be provided for this theorem. Each proof will consist of two parts.

- Part I: If $PQRS$ is a rectangle, then $PR≅QS .$
- Part II: If $PR≅QS ,$ then $PQRS$ is a rectangle.

This proof will use similar triangles to prove the theorem.

Suppose $PQRS$ is a rectangle and $PR$ and $QS $ are its diagonals. By the Parallelogram Opposite Sides Theorem, the opposite sides of a parallelogram are congruent. Therefore, $RS$ and $QP $ are congruent. Additionally, by the Reflexive Property of Congruence, $SP,$ or $PS,$ is congruent to itself.

Since the angles of a rectangle are right angles, by the definition of congruent angles, $∠RSP≅∠QPS$. Consequently, $△RSP$ and $△QPS$ have two pairs of congruent sides and congruent included angles.$RS≅QP ∠RSP≅∠QPSSP≅PS $

Therefore, by the Side-Angle-Side Congruence Theorem, the triangles are congruent.
$△RSP≅△QPS $

Because corresponding parts of congruent triangles are congruent, $PR$ and $QS ,$ which are the diagonals of $PQRS,$ are congruent.
$PR≅QS $

Consider the parallelogram $PQRS$ and its diagonals $PR$ and $QS $ such that $PR≅QS .$

By the Parallelogram Opposite Sides Theorem, $PQ ≅SR.$ Additionally, by the Reflexive Property of Congruence, $PS$ is congruent to itself.

The sides of $△QPS$ are congruent to the sides of $△RSP.$$SQ ≅PRSR≅PQ PS≅SP $

Therefore, by the Side-Side-Side Congruence Theorem, $△QPS≅△RSP.$ Moreover, since corresponding parts of congruent triangles are congruent, $∠QPS$ is congruent to $∠RSP.$
$∠QPS≅∠RSP⇕m∠QPS=m∠RSP $

Note that $∠QPS$ and $∠RSP$ are consecutive angles. By the Parallelogram Consecutive Angles Theorem, these angles are supplementary. With this information, it can be concluded that both $∠QPS$ and $∠RSP$ are right angles.
$m∠QPS+m∠RSP=180_{∘}⇓m∠QPS=90_{∘}andm∠RSP=90_{∘} $

Additionally, by the Parallelogram Opposite Angles Theorem, $∠QPS≅∠SRQ$ and $∠RSP≅PQR.$ Because all of the angles are right angles, $PQRS$ is a rectangle.
This proof will use transformations to prove the theorem.

Consider the rectangle $PQRS$ and its diagonals $PR$ and $QS .$ Let $M$ be the point of intersection of the diagonals.

Let $A$ and $B$ be the midpoints of $PS$ and $RQ .$ Then, a line through $M$ and the midpoints $A$ and $B$ can be drawn.

Note that $QA ,$ $AR,$ $PB,$ and $BS$ are congruent segments. Because congruent segments have the same length, the distance between $Q$ and $A$ equals the distance between $R$ and $A.$ Therefore, $Q$ is the image of $R$ after a reflection across $AB.$ Similarly, $P$ is the image of $S$ after the same reflection.
Since $M$ lies on $AB,$ a reflection across $AB$ maps $M$ onto itself.

Reflection Across $AB$ | |
---|---|

Preimage | Image |

$R$ | $Q$ |

$S$ | $P$ |

$M$ | $M$ |

Because corresponding parts of congruent figures are congruent, $QM ≅RM$ and $PM≅SM.$ Additionally, by the Parallelogram Diagonals Theorem, the diagonals of the rectangle bisect each other. Therefore, all four segments are congruent. Each diagonal of the parallelogram consists of the same two congruent segments. By the Segment Addition Postulate, the diagonals are congruent.

$PR≅QS $

Consider the parallelogram $PQRS$ and its diagonals $PR$ and $QS $ such that $PR≅QS .$ By the Parallelogram Diagonals Theorem, the diagonals of a rectangle bisect each other at $M.$

By the Parallelogram Opposite Sides Theorem, $PQ ≅SR$ and $QR ≅PS.$

Let $A$ and $B$ be the midpoints of $PS$ and $RQ .$ Then, a line through $M$ and the midpoints $A$ and $B$ can be drawn.

As shown before, $Q,$ $P,$ and $M$ are the respective images of $R,$ $S,$ and $M$ after a reflection across $AB.$ Therefore, since $△QPM$ is the image of $△RSM$ after a reflection across $AB,$ the triangles are congruent.
Let $C$ and $D$ be the midpoints of of $PQ $ and $RS.$ By following the same reasoning, $△PMS$ is the image of $△QMR$ after a reflection across $CD.$ Therefore, the triangles are congruent.

The parallelogram consists of four triangles in which the opposite triangles are congruent. Therefore, the corresponding angles of these triangles are congruent. Additionally, because all triangles are isosceles, the angles opposite congruent sides are congruent as well.
Each angle of the parallelogram is the sum of the same two congruent angles. Therefore, all angles of the parallelogram are congruent.

$∠P≅∠Q≅∠R≅∠S $

Moreover, by the Parallelogram Consecutive Angles Theorem, $∠P$ and $∠S$ are supplementary. With this information, it can be concluded that both angles are right triangles.
$∠P+∠S=180_{∘}⇓m∠P=90_{∘}andm∠S=90_{∘} $

Because all of the angles are congruent, the angles of the parallelogram are right triangles. Therefore, $PQRS$ is a rectangle.
Zosia arrives early to a Harry Styles concert! She notices something about the stage, so she uses a napkin as paper and draws a diagram. The stage is a rectangle that she labels as $ABCD.$

Find the lengths of its diagonals to help Zosia understand the stage that she will see her favorite artist sing on!{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">A<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.07153em;\">C<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["31"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.05017em;\">B<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.02778em;\">D<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["31"]}}

In a rectangle, the diagonals are congruent.

By the Rectangle Diagonals Theorem, the diagonals of a rectangle are congruent. This means that $AC$ and $BD$ have the same length. With this information, an equation in terms of $x$ can be written.
The value of $x$ was found. Next, this value can be substituted in any of the expressions for the length of a diagonal. In this case, $x=7$ will be arbitrarily substituted in the expression for $AC.$
It was found that the length of $AC$ is $31.$ Since the diagonals of a rectangle are congruent, the length of $BD$ is also $31.$

$4x+3=45−2x $

The above equation will be solved for $x.$
$4x+3=45−2x$

$x=7$

$AC=31andBD=31 $

As with rectangles, it can also be determined whether a parallelogram is a rhombus just by looking at its diagonals.

A parallelogram is a rhombus if and only if its diagonals are perpendicular.

Based on the diagram, the following relation holds true.

Parallelogram $ABCD$ is a rhombus $⇔$ $<$