Here are a few recommended readings before getting started with this lesson.
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.
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. Note that congruent angles have the same measure. Let $x_{∘}$ be the measure of $∠A$ and $∠C,$ and $y_{∘}$ 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 $ABCD.$
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!First, for simplicity, the value of $x$ will be found. After that, the values of $a$ and $b$ will be calculated.
$(II):$ $LHS+10=RHS+10$
$(I):$ $a=3b$
$(I):$ Multiply
$(I):$ $LHS−10=RHS−10$
$(I):$ $LHS−10b=RHS−10b$
$(I):$ $LHS/5=RHS/5$
$(II):$ $b=10$
$(II):$ Multiply
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!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.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.
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.Reflection Across $AB$ | |
---|---|
Preimage | Image |
$R$ | $Q$ |
$S$ | $P$ |
$M$ | $M$ |
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.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!In a rectangle, the diagonals are congruent.
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 $⇔$ $AC⊥BD$
This proof will be written in two parts.
A rhombus is a parallelogram with four congruent sides. By the Parallelogram Diagonals Theorem, it can be said that its diagonals bisect each other. Let Let $ABCD$ be a rhombus with $P$ at the midpoint of both diagonals.
Note that $AB$ is congruent to $AD$ and $DP$ is congruent to $BP.$ Additionally, by the Reflexive Property of Congruence, $AP$ is congruent to itself. Therefore, by the Side-Side-Side Congruence Theorem, $△APB$ is congruent to $△APD.$ $⎩⎪⎪⎨⎪⎪⎧ AB≅ADDP≅BPAP≅AP ⇒△APB≅△APD $ Because corresponding parts of congruent triangles are congruent, $∠APB$ and $∠APD$ are congruent angles. Furthermore, these angles form a linear pair, which means they are supplementary. With this information, it can be concluded that both $∠APB$ and $∠APD$ are right angles. ${∠APB≅∠APDm∠APB+m∠APD=180_{∘} ⇓m∠APB=90_{∘}andm∠APD=90_{∘} $ This implies that $AC$ is perpendicular to $BD.$ Therefore, the diagonals of a rhombus are perpendicular.
Parallelogram $ABCD$ is a rhombus $⇒$ $AC⊥BD$
Conversely, let $ABCD$ be a parallelogram whose diagonals are perpendicular.
By the Parallelogram Diagonals Theorem, the diagonals of the parallelogram bisect each other. If $P$ is the midpoint of both diagonals, then $AP$ and $CP$ are congruent.
Since $AC$ and $BD$ are perpendicular, $∠APB$ and $∠CPB$ measure $90_{∘}$ and thus are congruent angles. By the Reflexive Property of Congruence, $BP$ is congruent to itself. This means that two sides and their included angle are congruent. By the Side-Angle-Side Congruence Theorem, $△APB$ and $△CPB$ are congruent triangles. $⎩⎪⎪⎨⎪⎪⎧ AP≅CP∠APB≅∠CPBBP≅BP ⇓△APB≅△CPB $ Because corresponding parts of congruent figures are congruent, it can be said that $AB$ is congruent to $CB.$
Furthermore, by the Parallelogram Opposite Sides Theorem, $AB$ is congruent to $DC$ and $AD$ is congruent to $BC.$ By the Transitive Property of Congruence, it follows that all sides of the parallelogram are congruent.
This means that if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.
$AC⊥BD$ $⇒$ parallelogram $ABCD$ is a rhombus
Zosia is now listening to Dua Lipa at home. Staring at some of her album covers, Zosia decides to design a parallelogram as the background art for Dua's next cover! She has made a parallelogram $ABCD$ in which the diagonals are perpendicular. To make a unique design, she wants to be sure of the length of $AB.$
Help Zosia draw the perfect design by finding the length of $AB!$If the diagonals of a parallelogram are perpendicular, then the quadrilateral is a rhombus.
By using the theorems seen in this lesson, other properties can be derived. One of them is the Parallelogram Consecutive Angles Theorem.
Parallelogram Consecutive Angles Theorem |
The consecutive angles of a parallelogram are supplementary. |
Furthermore, the theorems seen in this lesson can be applied to different parallelograms in different contexts. Consider a square. By definition, all its angles are right angles, and all its sides are congruent. Therefore, a square is both a rectangle and a rhombus.
Therefore, by the Rectangle Diagonals Theorem and the Rhombus Diagonals Theorem, the diagonals of a square are congruent and perpendicular.