# Exercise E8.1 Quadrilaterals NCERT Solutions Class 9

Exercise 8.1

## Chapter 8 Ex.8.1 Question 1

The angles of quadrilateral are in the ratio \(3:5:9:13\). Find all the angles of the quadrilateral.

**Solution**

**Video Solution**

**What is known?**

The angles of quadrilateral are in the ratio \(3:5:9:13.\)

**What is unknown?**

All the angles of the quadrilateral.

**Reasoning:**

Sum of angles in any quadrilateral is \(360\) degree.

**Steps:**

Let the common ratio between the angles be *\(x\)*.

Therefore, the angles will be \(3x\), \(5x\), \(9x\), and \(13x\) respectively. according to the given ratio.

As the sum of all interior angles of a quadrilateral is \(360^\circ\) ,

\[\begin{align} ∴\;3 x+5 x+9 x+13 x &=360^{\circ}\\30 x&=360^{\circ} \\x&=12^{\circ}\end{align}\]

Hence, the angles are

\[\begin{align}3 x&=3 \times 12=36^{\circ} \\ 5 x&=5 \times 12=60^{\circ} \\ 9 x&=9 \times 12=108^{\circ} \\ 13 x&=13 \times 12=156^{\circ}\end{align}\]

## Chapter 8 Ex.8.1 Question 2

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

**Solution**

**Video Solution**

**What is known/?**

The diagonals of a parallelogram are equal.

**What is unknown?**

How we can show that it is a rectangle.

**Reasoning: **

To show that given parallelogram is a rectangle, we have to prove that one of its interior angles is and this can be done by showing two triangles congruent.

**Steps:**

Let \(ABCD\) be a parallelogram. To show that \(ABCD\) is a rectangle, we have to prove that one of its interior angles is \(\begin{align}90^{\circ}\end{align}\) .

In \(\Delta ABC\) and \(\Delta DCB\)

\[\begin{align} & AB=DC\, \\ & \left(\!\begin{array} & \text{In a parallelogram, } \\ \text{opposite sides are equal}\text{.} \\ \end{array}\!\right) \\ & \\ & BC=BC\,\left( \text{Common} \right) \\ & AC=DB\,\left( \text{Given} \right) \\ & \therefore \Delta ABC\,\cong \,\Delta DCB \\ & \left( By\,SSS\,\text{Congruence rule} \right) \\ & \\ & \Rightarrow \,\,\,\angle ABC=\angle DCB \\ \end{align}\]

It is known that the sum of the measures of angles on the same side of transversal is \(180^\circ\).

\[\begin{align}\angle &{\rm{ABC}} + \angle {\rm{DCB}} = {\rm{18}}{0^0}\;\left( {{\rm{AB }}||{\rm{ CD}}} \right)\\&\Rightarrow {\rm{ }}\angle {\rm{ABC}} + \angle {\rm{ABC}} = {\rm{18}}{0^0}\\&\Rightarrow {\rm{ 2}}\angle {\rm{ABC}} = {\rm{18}}{0^0}\\& \Rightarrow {\rm{ }}\angle {\rm{ABC}} = {\rm{9}}{0^0}{\rm{ }}\end{align}\]

Since \(ABCD\) is a parallelogram and one of its interior angles is \(\begin{align}90^{\circ},\end{align}\) \(ABCD\) is a rectangle.

## Chapter 8 Ex.8.1 Question 3

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

**Solution**

**Video Solution**

**What is known?**

The diagonals of a quadrilateral bisect each other at right angles.

**What is unknown?**

How we can show that it is a rhombus.

**Reasoning: **

To show that given quadrilateral is a rhombus, we have to show it is a parallelogram and all the sides are equal.

**Steps:**

Let be a quadrilateral, whose diagonals and bisect each other at right angle i.e

\(\mathrm{OA}\!=\!\mathrm{OC}, \mathrm{OB}\!=\!\mathrm{OD},\) and

\( \angle \mathrm{AOB}\!=\!\angle \mathrm{BOC}\!=\!\angle \mathrm{COD}\!=\!\angle \mathrm{AOD}\!=\!90^{\circ}\)

To prove \(ABCD\) a rhombus,

We have to prove \(ABCD\) is a parallelogram and all the sides of \(ABCD\) are equal.

In \(\Delta A O D \text { and } \Delta C O D,\)

\[\begin{align}O A&=O C\\(\text { Diagonals }&\text{bisect each other )} \\\\ \angle A O D&=\angle C O D \quad \text { (Given) } \\ O D&=O D \quad \text { (Common) } \\ ∴ \Delta A O D &\cong \Delta C O D \\ \text { (By SAS}&\text{ congruence rule) } \\\\ \therefore A D&=C D\quad\dots\text{(1)}\end{align}\]

Similarly, it can be proved that

\[\begin{align}A D=A B \text { and } C D=B C\quad\dots\text{(2)}\end{align}\]

From Equations (1) and (2),

\[\begin{align}A B=B C=C D=A D\end{align}\]

Since opposite sides of quadrilateral \(ABCD\) are equal, it can be said that \(ABCD\) is a parallelogram. Since all sides of a parallelogram \(ABCD\) are equal, it can be said that \(ABCD\) is a rhombus.

## Chapter 8 Ex.8.1 Question 4

Show that the diagonals of a square are equal and bisect each other at right angles.

**Solution**

**Video Solution**

**What is known?**

Given quadrilateral is a square.

**What is unknown?**

How we can show that diagonals of a square are equal and bisect each other at right angles.

**Reasoning: **

By showing two triangles congruent consist diagonals and then we can say corresponding parts of congruent triangles are equal.

**Steps:**

Let \(ABCD\) be a square. Let the diagonals \(AC\) and \(BD\) intersect each other at a point \(O\). To prove that the diagonals of a square are equal and bisect each other at right angles, we have to prove,

\(\begin{align}&\mathrm{AC}=\mathrm{BD}, \mathrm{OA}=\mathrm{OC}, \mathrm{OB}=\mathrm{OD},\end{align}\) and \(\begin{align} \angle \mathrm{AOB}=90^{\circ} \end{align}\)

In \(\triangle \mathrm{ABC} \text { and } \Delta \mathrm{DCB}\)

\[\begin{align} AB&={DC}\left[\begin{array}{}\text {Sides of a square are}\\ \text{equal to each other}\end{array}\right]\\{\angle {ABC}}&\!=\!\angle {DCB} \left[\!\begin{array}{}\text{All interior angles }\\ \text{ are of }90^\circ \end{array}\!\right]\\ {{BC}}&={{CB} \quad \text{(Common side) }} \\ ∴ \Delta {ABC} &\cong \Delta {DCB} \left[\begin{array}{}\text{By SAS }\\ \text{congruency}\end{array}\right] \\ {∴ \; {AC}}&={{DB} \quad\text{(By CPCT )}}\end{align}\]

Hence, the diagonals of a square are equal in length.

In \(\Delta {{AOB \;\rm and }}\;\Delta {{COD ,}}\)

\[\begin{align}\angle {{AOB}} &= \angle {{COD}}\\\text{(Vertically }&\text {opposite angles)}\\\\\angle {{ABO}} &= \angle {{CDO}}\\\text{(Alternate}&\text { interior angles)}\\\\{{AB}} &= {{CD}}\\\text{( Sides of a square}&\text { are always equal)}\\\\\therefore \Delta {{AOB}}&\cong \Delta {{COD }}\\ \text{(By AAS}&\text { congruence rule)}\\\\ \therefore {{AO}} &= {{CO \;\rm and \;OB}} = {{OD}}\\ &\text{(By CPCT)}\end{align}\]

Hence, the diagonals of a square bisect each other.

In \(\begin{align} \Delta {AOB} \end{align}\) and \(\begin{align} \Delta {COB}, \end{align}\)

As we had proved that diagonals bisect each other, therefore,

\[\begin{align} {{AO}}&={{CO}} \\ {{AB}}&={CB}\\ \text{( Sides of a}&\text { square are equal )} \\ \\{{BO}}&={{BO} \,\quad \text{( Common )}}\\{∴\Delta {AOB} }&\cong \Delta {COB}\\ \text{(By SSS}&\text { congruency )} \\\\ \therefore {\angle {AOB}}&={\angle {COB}\,\quad \text{(By CPCT)}}\end{align}\]

However,

\[\begin{align}\angle {AOB}+\angle {COB}&=180^{\circ} \\2 \angle {AOB} &=180^{\circ} \\ \angle {AOB}&=90^{\circ}\end{align}\]

Hence, the diagonals of a square bisect each other at right angles.

## Chapter 8 Ex.8.1 Question 5

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

**Solution**

**Video Solution**

**What is known?**

The diagonals of a quadrilateral are equal and bisect each other at right angle.

**What is unknown?**

How we can show that it is a square.

**Reasoning: **

We have to show that given quadrilateral is a parallelogram in which all sides are equal and one of its interior angles is .

**Steps:**

Let us consider a quadrilateral \(ABCD\) in which the diagonals \(AC\) and \(BD\) intersect each other at \(O\). It is given that the diagonals of \(ABCD\) are equal and bisect each other at right angles.

Therefore,

\({\rm{AC }} = {\rm{ BD}},{\rm{ OA }} = {\rm{ OC}},{\rm{ OB }} = {\rm{ OD}},\) and

\(\angle \,{\rm{AOB }} \!=\!\angle {\rm{BOC }} \!=\! \angle {\rm{COD }} \!=\! \angle {\rm{AOD }} \!=\! {\rm{ 9}}{0^0}\)

To prove \(ABCD\) is a square

We have to prove that \(ABCD\) is a parallelogram,in which

\(AB = BC = CD = AD\), and one of its interior angles is \(90^{0}\) .

In \( { \Delta{AOD} }{ {\: \rm and } \:\Delta {COD}}\)

\[\begin{align}{{AO}}&={CO}\\ \text{(Diagonals }& \text { bisect each other)} \\\\ \angle {AOD}&=\angle {COD}\\( \text{ Given that} &\text{ each is } 90^{\circ}) \\\\ {{OD}}&={{OD}\quad( \text{ Common })}\\ \therefore \Delta {AOD} &\cong \Delta {COD}\\ \text{(SAS}& \text { congruence rule)} \\\\ \therefore {AD}&={{DC}}\quad\dots(3)\end{align}\]

\[\rm {And,}\angle \mathrm{OAB}=\angle \mathrm{OCD}(\mathrm{By} \,\text {CPCT})\]

However, these are alternate interior angles for line \(AB\) and \(CD\) and alternate interior angles are equal to each other only when the two lines are parallel.

\(AB \;|| \;CD \)... (2)

From Equations (1) and (2), we obtain \(ABCD\) is a parallelogram.

In \(\Delta AOD\) and \(\Delta {COD}\)

\[\begin{align} {{AO}}&={CO}\\ \text{(Diagonals }& \text { bisect each other)} \\ \\\ {\angle {AOD}}&=\angle {COD}\\( \text{ Given that} &\text{ each is } 90^{\circ})\\\\ {{OD}}&={{OD}\,\,( \text{ Common })}\\ \therefore \Delta {AOD} &\cong \Delta {COD}\\ \text{(SAS}& \text { congruence rule)} \\\\ \therefore {AD}&={{DC}}\qquad\dots(3)\end{align}\]

However, \(AD = BC\) and \(AB = CD\)

(Opposite sides of parallelogram \(ABCD\))

\(AB = BC = CD = DA\)

Therefore, all the sides of quadrilateral \(ABCD\) are equal to each other.

In \( \Delta {ADC}\) and \( \Delta {BCD} ,\)

\[\begin{align}\\ {{AD}}&={{BC}\; ( \text{ Already proved })} \\ {{AC}}&={{BD} \;\text{ (Given) }} \\ {{DC}}&={{CD}\; ( \text{ Common })}\\{\therefore \Delta {ADC} }&\cong \Delta {BCD}\\ \text{(SAS}& \text { congruence rule)} \\\\ {\therefore\angle {ADC}}&={\angle {BCD}\;(\text{By} \,\text{CPCT})}\end{align}\]

However,

\[\begin{align} \angle {ADC}\!+\angle {BCD}&=180^{\circ}\\ \text{ (Co-interior }& \text {angles) } \\\\ {\qquad \angle {ADC}+\angle {ADC}}&={180^{\circ}}\\{2 \angle {ADC}}&={180^{\circ}} \\ {\therefore \angle {ADC}}&={90^{\circ}}\end{align}\]

One of the interior angles of quadrilateral \(ABCD\) is a right angle.

Thus, we have obtained that \(ABCD\) is a parallelogram, \(AB = BC = CD = AD\) and one of its interior angles is \(90^{\circ}\) .

Therefore, \(ABCD\) is a square.

## Chapter 8 Ex.8.1 Question 6

Diagonal \(AC\) of a parallelogram \(ABCD\) bisects \(\angle A\) (see the given figure). Show that

(i) It bisects \(\angle C\) also,

(ii) \(ABCD\) is a rhombus.

**Solution**

**Video Solution**

**What is known?**

The diagonal AC of a parallelogram \(ABCD\) bisects \(∠A.\)

**What is unknown?**

How we can show that (i) It bisects \(∠C\) also, (ii) ABCD is a rhombus.

**Reasoning: **

We can use alternate interior angles property to show diagonal \(AC\) bisects angle \(C\) also, by showing all sides equal it can be said rhombus.

**Steps:**

(i) \(ABCD\) is a parallelogram.

\(\begin{align} &\angle \mathrm{DAC}=\angle {BCA}\\&\text{(Alternate interior angles are equal)} \ldots(1)\end{align}\)

\(\begin{align}\text{And,} &\angle \mathrm{BAC}=\angle {DCA}\\&\text{(Alternate interior angles are equal)} \ldots(2)\end{align}\)

However, it is given that \(AC\) bisects \(\angle A\).

\(\angle \mathrm{DAC}=\angle \mathrm{BAC} \ldots(3)\)

From Equations (\(1\)),(\(2\)), and (\(3\)), we obtain

\[\begin{align} {\angle D A C}&\!=\!\!{\angle B C A\!=\!\!\angle B A C\!=\!\!\angle D C A \ldots\!(\!4\!)} \\ {\angle D C A}&\!=\!\!{\angle B C A}\end{align}\]

Hence, \(AC\) bisects \(\angle C\).

(ii) From Equation (\(4\)), we obtain

\(\angle {DAC}=\angle {DCA}\)

\({{DA}}={DC}\)(Side opposite to equal angles are equal)

However, \(DA = BC\) and \(AB = CD\)

(Opposite sides of a parallelogram)

\(AB = BC = CD = DA \)

Hence, \(ABCD\) is a rhombus.

## Chapter 8 Ex.8.1 Question 7

\(ABCD\) is a rhombus. Show that diagonal \(AC\) bisects \(\angle A\) as well as \(\angle C\) and diagonal \(BD\) bisects \(\angle B\) as well as \(\angle D\).

**Solution**

**Video Solution**

**What is known?**

ABCD is a rhombus.

**What is unknown?**

How we can show that diagonal \(AC\) bisects \(∠A\) as well as \(∠C\) and diagonal BD bisects \(∠B\) as well as \(∠D.\)

**Reasoning: **

We can use alternate interior angles property to show diagonal \(AC\) bisects angles \(A\) and \(C\), similarly diagonal \(BD\) bisects angles B and D.

**Steps:**

Let us join \(AC\).

In \(\Delta\)\(ABC\),

\[\begin{align}&BC=AB\\&\left( \begin{array} & \text{Sides of a rhombus are} \\ \text{ equal to each other} \\ \end{array} \right)\\\\&\angle1=\angle2\\&\left( \begin{array} & \text{Angles opposite to equal sides } \\ \text{ of a triangle are equal} \\

\end{array} \right)\end{align}\]

However, \(\angle 1 = \angle 3\) (Alternate interior angles for parallel lines \(AB\) and \(CD\))

\(\angle 2 = \angle 3\)

Therefore, \(AC\) bisects \(\angle C\).

Also, \(\angle 2 = \angle 4\) (Alternate interior angles for || lines \(BC\) and \(DA\)) \(\angle 1 = \angle 4\)

Therefore, \(AC\) bisects \(\angle A\).

Similarly, it can be proved that \(BD\) bisects \(\angle B\) and \(\angle D\) as well.

## Chapter 8 Ex.8.1 Question 8

\(ABCD\) is a rectangle in which diagonal \(AC\) bisects \(\angle A\)as well as \(\angle C\). Show that:

(i) \(ABCD\) is a square

(ii) Diagonal \(BD\) bisects \(\angle B\) as well as \(\angle D\).

**Solution**

**Video Solution**

**What is known?**

\(ABCD\) is a rectangle in which diagonal \(AC\) bisects \(∠A\) as well as \(∠C.\)

**What is unknown?**

How we can show that

(i) \(ABCD\) is a square

(ii) Diagonal \(BD\) bisects \(∠B\) as well as \(∠D.\)

**Reasoning: **

We can use angle bisector property and isosceles triangle property to show given rectangle as square and alternate interior angles property to show BD bisects angles B and D.

**Steps:**

(i) It is given that \(ABCD\) is a rectangle.

\[\begin{align}\angle {\rm{A}} &= \angle {\rm{C}}\\\Rightarrow \frac{1}{2}\angle {\rm{A}} &= \frac{1}{2}\angle {\rm{C}}\\\Rightarrow \angle \rm DAC &= \frac{1}{2}\angle {\rm{DCA }} \\({\text{AC bisects }}&\angle {\rm{A \,and }}\angle\,\, {\rm{C}})\end{align}\]

\(CD = DA \) (Sides opposite to equal angles are also equal)

However, \(DA = BC\) and \(AB = CD\)

(Opposite sides of a rectangle are equal)

\(AB = BC = CD = DA\)

\(ABCD\) is a rectangle and all the sides are equal.

Hence, \(ABCD\) is a square.

(ii) Let us join \(BD\).

In \(\Delta {BCD}\)

\(BC=CD\) (Sides of a square are equal to each other)

\(\angle CDB=\angle CBD \)(Angles opposite to equal sides are equal)

However, \(\angle CDB = \angle ABD\) (Alternate interior angles for \(AB \;||\; CD\))

\[\angle CDB = \angle ABD\]

\(BD\) bisects \(\angle B\)

Also, \(\angle CDB = \angle ABD\) (Alternate interior angles for \(BC \;||\; AD\)) \(\angle CDB = \angle ABD\)

\(BD\) bisects \(\angle D\) and \(\angle B\).

## Chapter 8 Ex.8.1 Question 9

In parallelogram \(ABCD\), two points \(P\) and \(Q\) are taken on diagonal \(BD\) such that

\(DP = BQ\) (see the given figure). Show that:

\(\begin{align}&{\text { (i) } \triangle \mathrm{APD} \cong \Delta \mathrm{CQB}} \\ &{\text { (ii) } \mathrm{AP}=\mathrm{CQ}} \\ &{\text { (iii) } \triangle \mathrm{AQB} \cong \Delta \mathrm{CPD}} \\ &{\text { (iv) } \mathrm{AQ}=\mathrm{CP}} \\ &{\text { (v) APCQ is a parallelogram }}\end{align}\)

**Solution**

**Video Solution**

**What is known?**

\(ABCD\) is a parallelogram and \(DP = BQ\)

**What is unknown?**

How we can show that

\(\begin{align} &\left(\text{i} \right)\text{ }\ \ \ \Delta APD\cong \Delta CQB \\ &\left( \text{ii} \right)\text{ }\ \ AP=CQ \\ &\left( \text{iii} \right)\text{ }\ \Delta AQB\cong \Delta CPD \\ &\left(\text{ iv} \right)\text{ }\ AQ=CP \\ &\left( \text{v} \right)\text{ }\ APCQ\text{ }is\text{ }a\text{ parallelogram} \end{align}\)

**Reasoning: **

We can use alternate interior angles and parallelogram property for triangles congruence criterion to show triangles congruent then we can say corresponding parts of congruent triangles will be equal.

**Steps:**

(i) \(In{\text{ }}\Delta APD\;{\text{and}}{\text{ }}\Delta CQB,\)

\[\begin{align} \angle \mathrm{ADP} &\!=\!\angle \mathrm{CBQ}\!\begin{bmatrix}\text {Alternate interior}\\ \text{angles for BC } \| \mathrm{AD}\!\end{bmatrix} \\ \mathrm{AD}&\!=\!\mathrm{CB}\begin{bmatrix}\text {Opposite sides of} \\\text{parallelogram } \mathrm{ABCD}\end{bmatrix} \\ \mathrm{DP} &=\mathrm{BQ}\;(\text {Given}) \\ \therefore\triangle \mathrm{APD} &\cong \Delta \mathrm{CQB}\;\begin{bmatrix}\text { Using SAS }\\\text{congruence rule }\end{bmatrix} \end{align}\]

(ii) As we had observed that

\(\Delta APD \cong \Delta CQB\)

\(\begin{align} \therefore\, \mathrm{AP}=\mathrm{CQ}(\mathrm{CPCT})\end{align}\)

(iii) In \(\Delta AQB\;{\text{and}}{\text{ }}\Delta CPD,\)

\[\begin{align}{\angle \mathrm{ABQ}}&\!=\!\angle \mathrm{CDP}\!\begin{bmatrix}\! \text {Alternate interior}\\\text{ angles for } \mathrm{AB} \| \mathrm{CD} \!\end{bmatrix} \\ {\mathrm{AB}}&\!=\!\mathrm{CD}\begin{bmatrix}\text {Opposite sides of} \\\text{parallelogram } \mathrm{ABCD}\end{bmatrix} \\ {\mathrm{BQ}}&={\mathrm{DP}\;(\text {Given})} \\ {\therefore \Delta \mathrm{AQB} }&\cong \triangle \mathrm{CPD}\;\begin{bmatrix}\text { Using SAS }\\\text{congruence rule }\end{bmatrix} \end{align}\]

(iv) As we had observed that

\(\Delta AQB \cong \Delta CPD,\)

\(\begin{align}\therefore \mathrm{AQ}=\mathrm{CP}(\mathrm{CPCT})\end{align}\)

(v) From the result obtained in (ii) and (iv),

\[\begin{align} \text { AQ }&=\mathrm{CP} \text { and } \\ {\mathrm{AP}}&={\mathrm{CQ}}\end{align}\]

Since opposite sides in quadrilateral \(APCQ\) are equal to each other, \(APCQ\) is a parallelogram.

## Chapter 8 Ex.8.1 Question 10

\(ABCD\) is a parallelogram and \(AP\) and \(CQ\) are perpendiculars from vertices \(A\) and \(C\) on

diagonal \(BD\) (See the given figure). Show that

(i) \(\begin{align} &{ \Delta \mathrm{APB} }{\cong \Delta \mathrm{CQD}} \end{align}\)

(ii) \(\begin{align} &{ \mathrm{AP}=\mathrm{CQ}}\end{align}\)

**Solution**

**Video Solution**

**What is known?**

\(ABCD\) is a parallelogram and \(AP\bot DB,\,CQ\bot DB\)

**What is unknown?**

How we can show that (i) \(\Delta APB\cong \Delta CQD\)

(ii) \(AP = CQ\)

**Reasoning: **

We can use alternate interior angles and parallelogram property for triangles congruence criterion to show triangles congruent then we can say corresponding parts of congruent triangles will be equal.

**Steps:**

(i) In \(\begin{align} \;&\Delta \mathrm{APB} \text { and } \Delta \mathrm{CQD} \end{align}\)

\[\begin{align}\angle \mathrm{APB}&=\angle \mathrm{CQD}\begin{bmatrix} \text {Each angle} \\ \text{measures } 90^{\circ} \end{bmatrix} \\ {AB}&\!=\!{CD} \begin{bmatrix}\text {Opposite sides of} \\\text{parallelogram } {ABCD}\end{bmatrix} \\ \angle \mathrm{ABP}&\!=\!\angle \mathrm{CDQ} \begin{bmatrix}\text {Alternate interior}\\ \text{angles for AB } \| \mathrm{CD}\!\end{bmatrix} \\ \therefore \Delta \mathrm{APB} &\cong \Delta \mathrm{CQD}\begin{bmatrix}\text { By AAS }\\\text{congruence rule }\end{bmatrix} \end{align}\]

(ii) By using the above result

\(\begin{align} \triangle \mathrm{APB} \cong \Delta \mathrm{CQD}, \end{align}\) we obtain \(\begin{align} \mathrm{AP}=\mathrm{CQ}\end{align}\) (By CPCT )

## Chapter 8 Ex.8.1 Question 11

In \(\Delta ABC \,\rm{and} \,\Delta DEF\), \(AB \;|| \;DE\),

\(BC = EF\) and \(BC\: || \:EF\). Vertices \(A\), \(B\) and \(C\) are joined to vertices \(D\), \(E\) and \(F\) respectively (see the given figure). Show that

(i) Quadrilateral \(ABED\) is a parallelogram

(ii) Quadrilateral \(BEFC\) is a parallelogram

(iii) \(\begin{align} &{ \mathrm{AD} \| \mathrm{CF} \text { and } \mathrm{AD}=\mathrm{CF}} \end{align}\)

(iv) Quadrilateral \(ACFD\) is a parallelogram

(v) \(AC = DF\)

(vi) \(\begin{align} &{ \triangle \mathrm{ABC} \cong \triangle \mathrm{DEF} \text { . }}\end{align}\)

**Solution**

**Video Solution**

**What is known?**

In \(\Delta ABC\text{ }and\text{ }\Delta DEF,\)

\(AB = DE, AB || DE, BC = EF\) and \(BC || EF.\)

**What is unknown?**

How we can show that

(i) Quadrilateral \(ABED\) is a parallelogram

(ii) Quadrilateral \(BEFC\) is a parallelogram

(iii) \(AD || CF\) and \(AD = CF\)

(iv) Quadrilateral \(ACFD\) is a parallelogram

(v) \(AC = DF\)

(vi) \(\Delta ABC\cong \Delta DEF.\)

**Reasoning: **

We can use the fact that in a quadrilateral if one pair of opposite sides are parallel and equal to each other then it will be a parallelogram and converse is also true. Also by using suitable congruence criterion we can show triangles congruent then we can say corresponding parts of congruent triangles will be equal.

**Steps:**

(i) It is given that \(AB = DE\) and \(AB \;|| \;DE\).

If one pair of opposite sides of a quadrilateral are equal and parallel to each other, then it will be a parallelogram.

Therefore, quadrilateral \(ABED\) is a

parallelogram.

(ii) Again, \(BC = EF\) and \(BC\: || \:EF\)

Therefore, quadrilateral \(BCFE\)

is a parallelogram.

(iii) As we had observed that \(ABED\) and \(BEFC\) are parallelograms, therefore

\(AD = BE\) and \(AD \;|| \;BE\)

(Opposite sides of a parallelogram are equal and parallel)

And, \(BE = CF\) and \(BE\; || \;CF\)

(Opposite sides of a parallelogram are equal and parallel)

\(\therefore\) \(AD = CF\) and \(AD \;|| \;CF\)

(iv) As we had observed that one pair of opposite sides (\(AD\) and \(CF\)) of quadrilateral \(ACFD\) are equal and parallel to each other, therefore, it is a parallelogram.

(v) As \(ACFD\) is a parallelogram, therefore, the pair of opposite sides will be equal and parallel to each other.

\(\therefore\) \(AC\; || \;DF\) and \(AC = DF\)

(vi) \(\begin{align} { \triangle \mathrm{ABC} \text { and } \triangle \mathrm{DEF} \text { , }} \end{align}\)

\[\begin{align} \mathrm{AB} &=\mathrm{DE}(\text {Given}) \\ \mathrm{BC} &=\mathrm{EF}(\text { Given}) \\ \mathrm{AC} &=\mathrm{DF}\\(\mathrm{ACFD} &\text { is a parallelogram) }\\ \therefore \triangle \mathrm{ABC} &\cong \Delta \mathrm{DEF}\\\text{(By SSS} &\text { congruence rule })\end{align}\]

## Chapter 8 Ex.8.1 Question 12

\(ABCD\) is a trapezium in which \(AB\; ||\; CD\) and \(AD = BC\) (see the given figure). Show that

(i) \(\begin{align} &{ \angle {A}=\angle {B}} \end{align}\)

(ii) \(\begin{align} & { \angle {C}=\angle {D}} \end{align}\)

(iii) \(\begin{align}&{ \triangle {ABC} \cong \triangle {BAD}} \end{align}\)

(iv) diagonal \(AC =\) diagonal \(BD\)

[*Hint*: Extend \(AB\) and draw a line through \(C\) parallel to \(DA\) intersecting \(AB\) produced at \(E\).]

**Solution**

**Video Solution**

**What is known?**

\(ABCD\) is a trapezium in which \(AB || CD\) and \(AD = BC.\)

**What is unknown?**

How we can show that

(i) \(∠A = ∠B\)

(ii) \(∠C = ∠D\)

(iii) \(∆ABC ≅ ∆BAD\)

(iv) diagonal \(AC =\) diagonal \(BD\)

**Reasoning: **

Consider parallel lines \(AD\) and \(CE, AE\) is the transversal line for them then sum of co-interior angles will be \(180\) degree also angles \(CBE\) and \(CBA\) are linear pairs gives sum \(180\), angles \(CEB\) equals to \(CBE\) because opposite to equal sides in triangle \(BCE\). Using these observations angle A is equal to angle B. Then AB is parallel to CD will help to show angle C is equal to angle D. Also, by using suitable congruence criterion we can show triangles congruent then we can say corresponding parts of congruent triangles will be equal.

**Steps:**

Let us extend \(AB\). Then, draw a line through \(C\), which is parallel to \(AD\), intersecting \(AB\) at point \(E\). It is clear that \(AECD\) is a parallelogram.

(i) \(AD = CE\) (Opposite sides of parallelogram \(AECD\))

However, \({AD}={BC}(\text { Given })\)

Therefore, \({BC}={CE}\)

\[\begin{align}&\angle {{CEB}} = \angle {{CBE}}\\&\left( \begin{array}{l}{\text{Angle opposite to equal }}\\{\text{sides are also equal}}\end{array} \right)\end{align}\]

Consider parallel lines and . is the transversal line for them.

\[\begin{align}&\angle {{A}} + \angle {{CEB}} = {180^\circ }\\&\left( \begin{array}{l}{\text{Angles on the same }}\\{\text{side of transversal}}\end{array} \right)\end{align}\]

\[\begin{align}&\angle {{A}} + \angle {{CBE}} = {180^\circ }\\&\left( \begin{array}{l}{\text{Using the relation }}\\\angle {{CEB}} = \angle {{CBE}}\end{array} \right) \ldots (1)\end{align}\]

However, \(\angle B\) + \(\angle CBE\) = (Linear pair angles) ... (2)

From Equations (1) and (2), we obtain

\(\angle A\) = \(\angle B\)

(ii)

\[\begin{align}&\angle A + \angle D = {180^\circ }\\&\left( \begin{array}{l}{\text{Angles on the same side}}\\{\text{ of the transversal}}\end{array} \right)\end{align}\]

Also,

\[\begin{align}&\angle C + \angle B = {180^\circ }\\&\left( \begin{array}{l}{\text{Angles on the same side }}\\{\text{of the transversal}}\end{array}\right)\end{align}\]

\[\therefore \angle A + \angle D = \angle C + \angle B\]

However,

\[\begin{align}&\angle A = \angle B\\&\left[ \begin{array}{l}{\text{Using the result }}\\{\text{obtained in (i) }}\end{array} \right]\\&\therefore \,\,\,\angle C = \angle D\end{align}\]

(iii) In \(\begin{align}\triangle {ABC} \text { and } \triangle {BAD}\end{align}\)

\[\begin{align}{{AB}} &= {{BA \text{(Common side) }}}\\{{BC}}& = AD\text {(Given) }\\\angle {{B}} &= \angle A {\text{(Proved before) }}\\\therefore \Delta {{ABC}} &\cong \Delta {{BAD }}\\&\left( {{\text{SAS congruence rule}}} \right)\end{align}\]

(iv) We had observed that,

\[\begin{align} \Delta {ABC}& \cong \Delta {BAD} \\ \;\;\;\;\therefore {AC}&= {{BD}({By} \;{CPCT})}\end{align}\]