NCERT Solutions for Exercise 8.1 Class 9 Maths Chapter 8 - Quadrilaterals Exercise 8.1

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

Ans: Given that the diagonals of a parallelogram are equal. Let us assume a parallelogram $ABCD$ whose diagonals are equal as shown in the figure below.

Here $ABCD$ is a parallelogram, so we can write that $AB=DC$ since opposite sides of a parallelogram are equal.

Consider the triangles $ABC$ and $DCB$. In these two triangles we have $AB=DC$, $BC=BC$ and as per given data $AC=DB$. So, we can write that $\Delta ABC\cong \Delta DCB$ by SSS Congruence rule.

$\Rightarrow \angle ABC=\angle DCB$

We know that the sum of the angles on the same side of the transversal is $180{}^\circ $.

$\therefore \angle ABC+\angle DCB=180{}^\circ $

Substitute the value $\angle ABC=\angle DCB$ in the above equation.

$\Rightarrow \angle ABC+\angle ABC=180{}^\circ $

$\Rightarrow 2\angle ABC=180{}^\circ $

$\Rightarrow \angle ABC=\dfrac{180{}^\circ }{2}$

$\Rightarrow \angle ABC=90{}^\circ $

Here we have the angle $ABC$ and $DCB$ as $90{}^\circ $. When the angles of the parallelogram are equal to $90{}^\circ $, then it is called as Rectangle.

Hence, if the diagonals of the parallelogram are equal then it is called as Rectangle.

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

Ans: Consider a square $ABCD$ whose diagonals are $AC$ and $BD$. The intersecting point of the diagonals will be $O$ as shown in below figure.

According to the given statement we need to prove that $AC=BD$, $OA=OC$, $OB=OD$ and $\angle AOB=90{}^\circ $.

Consider $\Delta ABC$ and $\Delta DCB$.

In both the triangles we can write 

$AB=DC$ since the sides of a square are equal to each other, $\angle ABC=\angle DCB=90{}^\circ $ and 

$BC=CB$.

So, by using SAS congruency law we can say that $\Delta ABC\cong \Delta DCB$.

$\therefore AC=DB$ by CPCT.

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

Now consider $\Delta AOB$ and $\Delta COD$.

In both the triangles we can write 

$\angle AOB=\angle COD$ since vertically opposite angles are equal, 

$\angle ABO=\angle CDO$ since alternate interior angles are also equal and 

$AB=CD$ since the sides of a square are always equal.

So, by using AAS congruence rule we can say that $\Delta AOB\cong \Delta COD$.

$\therefore AO=CO$ and $OB=OD$ by CPCT.

Now consider $\Delta AOB$ and $\Delta COB$,

As we have proved earlier $AO=CO$,

$AB=CB$ since sides of a square are equal,

$BO=BO$.

By using SSS congruency rule we can write that $\Delta AOB\cong \Delta COB$.

$\therefore \angle AOB=\angle COB$ by CPCT.

But the angles $AOB$ and $COB$ are linear pair.

$\therefore \angle AOB+\angle COB=180{}^\circ $

$2\angle AOB=180{}^\circ $

$\angle AOB=90{}^\circ $

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

3. Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle A$ as shown in figure. 

• Show that it bisects $\angle C$.

Ans: Given that $ABCD$ is a parallelogram.

$\therefore \angle DAC=\angle BCA$ and $\angle BAC=\angle DCA$ since alternative interior angles

But in the problem, we have given that $AC$ bisect $\angle A$.

$\therefore \angle DAC=\angle BAC$

From all the above equations we can write that 

$\angle DAC=\angle BCA=\angle BAC=\angle DCA$

$\angle DCA=\angle BAC$

Hence, $AC$ bisects $\angle C$.

• Show that $ABCD$ is a rhombus.

Ans: From the equation $\angle DAC=\angle BCA=\angle BAC=\angle DCA$, we can write that $\angle DAC=\angle DCA$.

$DA=DC$ since side opposite to equal angles are equal.

But we have $DA=BC$ and $AB=CD$ since opposite sides of the parallelogram are equal.

$\therefore AB=BC=CD=DA$

4. $ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ as well as $\angle C$. 

• $ABCD$ is a square.

Ans: Let the diagram of the given $ABCD$ rectangle is given by 

We have given that $ABCD$ is a rectangle. So, we can write that 

$\angle A=\angle C$

$\Rightarrow \dfrac{1}{2}\angle A=\dfrac{1}{2}\angle C$

But they have mentioned that $AC$ bisects $\angle A$ as well as $\angle C$.

$\Rightarrow \angle DAC=\dfrac{1}{2}\angle DCA$

$\therefore CD=DA$ since sides opposite to equal angles are also equal.

But $DA=BC$ and $AB=CD$ since opposite sides of a rectangle are equal.

$\therefore AB=BC=CD=DA$

Here $ABCD$ is given to be a rectangle but all the sides are equal.

Hence $ABCD$ is a square.

• Diagonal $BD$ bisects $\angle B$ as well as $\angle D$.

Ans: Now consider $\Delta BCD$, in this triangle we can write that 

$BC=CD$ since sides of a square are equal,

$\angle CDB=\angle CBD$ since angles opposite to equal sides are equal.

But we have $\angle CDB=\angle ADB$ since alternate interior angles for parallel lines $AB$ and $CD$.

$\therefore \angle CBD=\angle ABD$

Hence $BD$ bisects $\angle B$.

Also, $\angle CBD=\angle ADB$ which are alternate interior angles for parallel lines $BC$ and $AD$.

$\therefore \angle CDB=\angle ADB$

Finally, $BD$ bisects $\angle D$ and $\angle B$.

5. In parallelogram $ABCD$, two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP=BQ$ as shown in below figure. 

• Show that $\Delta APD\cong \Delta CQB$

Ans: Consider $\Delta APD$ and $\Delta CQB$, in these two triangles we can write 

$\angle ADP=\angle CBQ$ since alternate interior angles for the parallel lines $BC$ and $AD$,

$AD=CB$ since opposite sides of parallelogram are equal.

$DP=BQ$ which is in the given data.

By using SAS congruence rule, we can say that $\Delta APD\cong \Delta CQB$

• Show that $AP=CQ$

Ans: As we have $\Delta APD\cong \Delta CQB$. By using CPCT we can write that $AP=CQ$.

• Show that $\Delta AQB\cong \Delta CPD$

Ans: Consider $\Delta AQB$ and $\Delta CPD$, in these two triangles we can write 

$\angle ABQ=\angle CDP$ since alternate interior angles for the parallel lines $AB$ and $CD$,

$AB=CD$ since opposite sides of parallelogram are equal.

$BQ=DP$ which is in the given data.

By using SAS congruence rule, we can say that $\Delta AQB\cong \Delta CPD$

• Show that $AQ=CP$

Ans: As we have $\Delta AQB\cong \Delta CPD$. By using CPCT we can write that $AQ=CP$.

• Show that $APCQ$ is a Parallelogram.

Ans: From the results obtained in (ii) and (iv), which are $AQ=CP$ and $AP=CQ$. That means the opposite sides of the quadrilateral $APCQ$ are equal, so $APCQ$ is a parallelogram.

6. $ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$ as shown in figure. 

• Show that  $\Delta APB\cong \Delta CQD$

Ans: Consider $\Delta APB$ and $\Delta CQD$, in these two triangles we can write that 

$\angle APB=\angle CQD=90{}^\circ $,

$AB=CD$ since opposite sides of parallelogram,

$\angle ABP=\angle CDQ$ since alternate interior angles for parallel lines $AB$ and $CD$.

By using AAS congruency rule we can say that $\Delta APB\cong \Delta CQD$.

• Show that $AP=CQ$

Ans: From the statement $\Delta APB\cong \Delta CQD$, by using CPCT we can write that $AP=CQ$.

7. $ABCD$ is a trapezium on which $AB\parallel CD$ and $AD=BC$ as shown in figure. (Hint: Extend $AB$ and draw a line through $C$ parallel to $DA$ intersecting $AB$ at point $E$).

• Show that $\angle A=\angle B$

Ans: We have $AD=CE$ opposite sides of parallelogram and $AD=BC$.

$\therefore BC=CE$ and $\angle CEB=\angle CBE$ since angle opposite to equal sides are also equal.

Consider the parallel lines $AD$ and $CE$. $AE$ is the transversal line for them.

So, 

$\angle A+\angle CEB=180{}^\circ $ since angles on the same side of transversal.

$\Rightarrow \angle A+\angle CBE=180{}^\circ $

Also, $\angle B+\angle CBE=180{}^\circ $ as they are linear pair of angles. So, from these two equals we can write that 

$\angle A=\angle B$

• Show that $\angle C=\angle D$

Ans: Given $AB\parallel CD$, so 

$\angle A+\angle D=180{}^\circ $ , $\angle C+\angle B=180{}^\circ $since angles on the same side of the transversal.

$\therefore \angle A+\angle D=\angle C+\angle B$

But we have $\angle A=\angle B$, so we will have 

$\angle C=\angle D$

• Show that $\Delta ABC\cong \Delta BAD$

Ans: Consider $\Delta ABC$ and $\Delta BAD$, in these two triangles we can write that 

$AB=BA$, $BC=AD$ and $\angle B=\angle A$.

By using SAS congruence rule, we can say that $\Delta ABC\cong \Delta BAD$.

• Show that diagonal $AC=\text{ diagonal }BD$

Ans: We have $\Delta ABC\cong \Delta BAD$, so by using CPCT we can write that $AC=BD$

Conclusion

NCERT Class 9 Maths Ex 8.1 is a crucial exercise that teaches students the properties and classifications of quadrilaterals, including squares, rectangles, rhombuses, and parallelograms. It builds foundational understanding for more complex geometrical concepts and enhances analytical skills by asking students to prove or disprove certain characteristics of these shapes. Mastering quadrilaterals class 9 exercise 8.1 solutions can help students secure marks and deepen their understanding of geometry, as questions from this chapter have appeared frequently in exams.

Class 9 Maths Chapter 8 : Exercises Breakdown

CBSE Class 9 Maths Chapter 8 Other Study Materials

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for class 9th maths chapter 8 exercise 8.1 Go through these chapter-wise solutions to be thoroughly familiar with the concepts.