NCERT Exemplar for Class 9 Maths Chapter 7 - Triangles - Free PDF Download

Sample Question 1: If $$\mathrm{\Delta }ABC\cong \mathrm{\Delta }PQR$$ and $$\mathrm{\Delta }ABC$$ is not congruent to $$\mathrm{\Delta }RPQ$$, then which of the following is not true: 

(A) $$BC=PQ$$

(B) $$AC=PR$$

(C) $$QR=BC$$

(D) $$AB=PQ$$

Ans: Correct option - A

Given, $$ABC\cong PQR$$

We know that:

Two triangles are said to be congruent if pairs of their corresponding sides and their corresponding angles are equal. 

Hence,

  $\text{AC}\text{ = }\text{PR}$

  $\text{AB}\text{ = }\text{PQ}$

  $\text{QR}\text{ = }\text{BC}$ 

Therefore, $$\text{BC}\text{ = }\text{PQ}$$ is not true for triangles.

EXERCISE 7.1

1. Which of the following is not a criterion for congruence of triangles?

A. SAS

B. ASA

C. SSA

D. SSS

Ans: Correct option - C

A. SAS: If two pairs of corresponding sides of two triangles are equal in length, and the corresponding included angles are equal in measurement, then the triangles are congruent.

B. ASA: If two pairs of corresponding angles of two triangles are equal in measurement, and the corresponding included sides are equal in length, then the triangles are congruent.

C. SSA: The SSA condition (Side-Side-Angle) which specifies two sides and a non-included angle (also known as ASS, or Angle-Side-Side) do not by itself prove congruence.

D. SSS: If three pairs of corresponding sides of two triangles are equal in length, then the triangles are congruent.

2. If $$AB=QR,BC=PR$$ and $$CA=PQ$$, then

A. $$\mathrm{\Delta }ABC\cong \mathrm{\Delta }PQR$$

B. $$\mathrm{\Delta }CBA\cong \mathrm{\Delta }PRQ$$

C. $$\mathrm{\Delta }BAC\cong \mathrm{\Delta }RPQ$$ 

D. $$\mathrm{\Delta }PQR\cong \mathrm{\Delta }BCA$$

Ans: Correct option - B

Given: $$\text{AB}\text{ = }\text{QR, BC}\text{ = }\text{PR and CA}\text{ = }\text{PQ}$$

$$\text{AB}\text{ = }\text{QR, BC}\text{ = }\text{PR and CA}\text{ = }\text{PQ}$$

Here, the three pairs of corresponding sides of two triangles are equal in length.

Apply the SSS rule of congruence, we have:

$$\mathrm{\Delta }CBA\cong \mathrm{\Delta }PQR$$  

3. In $$\mathrm{\Delta }ABC,AB=AC\text{ and }\mathrm{\angle }B={50}^{\circ }$$. Then $$\mathrm{\angle }C$$ is equal to

A. $40^{\circ }$

B. $50^{\circ }$

C. $80^{\circ }$  

D. $130^{\circ }$

Ans: Correct option - B

Given: $$\mathrm{\Delta }ABC,AB=AC\text{ and }\mathrm{\angle }B={50}^{\circ }$$

Since, $$AB=AC\text{ and }\mathrm{\angle }B={50}^{\circ }$$
$$\mathrm{\angle }B=\mathrm{\angle }C={50}^{\circ }$$          (Angles opposite to equal sides are equal)

4. In $$\mathrm{\Delta }ABC,BC=AB$$ and $$\mathrm{\angle }B={80}^{\circ }.$$ Then $$\mathrm{\angle }A$$ is equal to

A. $80^{\circ }$  

B. $40^{\circ }$

C. $50^{\circ }$ 

D. $100^{\circ }$

Ans: Correct option - C

The angles opposite to the equal side are equal.

$$BC=AB\text{ and }\mathrm{\angle }B={80}^{\circ }$$

let $$\mathrm{\angle }A$$ = $$x$$

$$\mathrm{\angle }A=\mathrm{\angle }C=x$$                             (As angles opposite to equal side are equal)

Now by angle sum property, we have:

 $\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$

  $\Rightarrow x+{80}^{\circ }+x={180}^{\circ }$

  $\Rightarrow 2x+{80}^{\circ }={180}^{\circ }$

  $\Rightarrow 2x={180}^{\circ }$

  $\Rightarrow x={\displaystyle \frac{{100}^{\circ }}{2}}={50}^{\circ }$ 

Therefore, $$x={50}^{\circ }$$

5. In $$\mathrm{\Delta }PQR,\mathrm{\angle }R=\mathrm{\angle }P$$ and QR = 4cm and PR = 5cm. 

Then the length of $$\text{PQ}$$ is

A. 4cm

B. 5cm

C. 2cm

D. 2.5cm 

Ans: Correct option - A

$$\mathrm{\angle }R=\mathrm{\angle }P\text{ and }QR=4\text{ cm and }PR=5\text{ cm}$$

Since, $$\mathrm{\angle }R=\mathrm{\angle }P$$

$$PQ=QR=4\text{cm}$$(Sides opposite to equal angles are equal)

6. $$D$$ is a point on the side $$BC$$ of a $$\mathrm{\Delta }ABC$$ such that $$AD$$ bisects $$\mathrm{\angle }BAC$$. 

Then,

A. BD  =  CD

B. BA  >  BD

C. BD  >  BA

D. CD  >  CA 

Ans: Correct option - B

Given, $\mathrm{\Delta }ABC$ such that $\text{AD}$ bisects $$\mathrm{\angle }BAC$$

The bisector divides a given angle into two angles with equal measures.

$\therefore \mathrm{\angle }BAD=\mathrm{\angle }CAD...\left(i\right)$

In $\mathrm{\Delta }ACD,\mathrm{\angle }BDA$ is an exterior angle.

$\therefore \mathrm{\angle }BDA>\mathrm{\angle }CAD...\left(ii\right)$            $[\because$ Exterior angle > interior opposite angle]

$\Rightarrow \mathrm{\angle }BDA>\mathrm{\angle }BAD$           {from Eq. (i)}

$\Rightarrow BA>BD\left[\text{side opposite to greater angle is greater}\right]$

7. It is given that $$\mathrm{\Delta }ABC\cong \mathrm{\Delta }FDE$$ and AB = 5cm,$$\mathrm{\angle }B={40}^{\circ }\text{ and }\mathrm{\angle }A={80}^{\circ }.$$ 

Then which of the following is true?

A. $DF=5cm,\mathrm{\angle }F={60}^{\circ }$

B. $DF=5cm,\mathrm{\angle }E={60}^{\circ }$

C. $DE=5cm,\mathrm{\angle }E={60}^{\circ }$

D. $DE=5cm,\mathrm{\angle }D={40}^{\circ }$ 

Ans: Correct option - B

It is given that $$\mathrm{\Delta }ABC\cong \mathrm{\Delta }FDE\text{ and }AB=5\text{cm},\mathrm{\angle }B={40}^{\circ }\text{ and}\mathrm{\angle }A={80}^{\circ }.$$

Two triangles are congruent if their corresponding sides are equal in length and their corresponding angles are equal in size.

In the first option, $$DF=5\text{cm}$$ but $$\mathrm{\angle }F$$ is not equal to $${60}^{\circ }.$$

In the second option, $$DF=5\text{cm}$$ and $$\mathrm{\angle }E={60}^{\circ }$$ is satisfied. 

In the third and the fourth options, the length of $$DE\ne 5\text{cm}$$, since it is not defined.

8. Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle cannot be

A. 3.6 cm

B. 4.1 cm 

C. 3.8 cm 

D. 3.4 cm 

Ans: Correct option - D

Two sides of a triangle are of lengths 5 cm and 1.5 cm.

In a triangle, the difference between two sides should be less than the third side.

In a triangle, the difference between two sides should be less than the third side.

Hence, option D is correct $$3.4\text{ cm}$$

9. In $$\mathrm{\Delta }PQR$$, if $$\mathrm{\angle }R>\mathrm{\angle }Q$$, then

A. QR  >  PR 

B. PQ  >  PR 

C. PQ  <  PR 

D. QR  <  PR 

Ans: Correct option - B

We have:

In $$\mathrm{\Delta }PQR$$, $$\mathrm{\angle }R>\mathrm{\angle }Q$$

We know that: 

The sides opposite greater angles are greater.

Hence, 

$$PQ>PR$$

10. In triangles $$ABC$$ and $$PQR,AB=AC,\mathrm{\angle }C=\mathrm{\angle }P$$ and $$\mathrm{\angle }B=\mathrm{\angle }Q$$. The two triangles are

A. isosceles but not congruent 

B. isosceles and congruent 

C. congruent but not isosceles 

D. neither congruent nor isosceles

Ans: Correct option - A

In $$\mathrm{\Delta }ABC\text{ and}\mathrm{\Delta }PQR$$

$$\mathrm{\angle }C=\mathrm{\angle }P$$(Given)

$$\mathrm{\angle }B=\mathrm{\angle }Q$$(Given)

$$\mathrm{\angle }A=\mathrm{\angle }R$$(Third angle of the triangle)

Also, given, $$AB=AC$$

Thus, $$\mathrm{\angle }B=\mathrm{\angle }C$$(Isosceles triangle Property)

But,$$\mathrm{\angle }B=\mathrm{\angle }Q$$and $$\mathrm{\angle }C=\mathrm{\angle }P$$

Hence, $$\mathrm{\angle }Q=\mathrm{\angle }P$$

$$PR=QR$$      (sides opposite equal angles are equal)

Thus, both the triangles are isosceles but not congruent.

11. In triangles $$ABC$$ and $$DEF,AB=FD$$ and $$\mathrm{\angle }A=\mathrm{\angle }D.$$ The two triangles will be congruent by SAS axiom if

A. BC = EF 

B. AC = DE 

C. AC = EF 

D. BC = DE 

Ans: Correct option - B

In triangles $$ABC\text{ and }DEF,AB=FD\text{ and}\mathrm{\angle }A=\mathrm{\angle }D.$$

SAS axiom:

If one angle between two adjacent sides of a triangle is similar to angle between two sides of another triangle then two triangles are congruent.

The two triangles $$ABC\text{ and }DEF$$ will be congruent by SAS axiom if $$AC=DE$$

Sample Question 1: In the two triangles $$ABC$$ and $$DEF,AB=DE$$ and $$AC=EF$$. Name two angles from the two triangles that must be equal so that the two triangles are congruent. Give reason for your answer. 

Ans: In the two triangles$$\text{ABC}$$$$\text{ and DEF},\text{ AB}\text{ = }\text{DE and AC}\text{ = }\text{EF}$$.

By SAS criterion of congruence, we have:

If one angle between two adjacent sides of a triangle is similar to angle between two sides of another triangle then two triangles are congruent. 

$$\mathrm{\Delta }ABC\cong \mathrm{\Delta }EDF\text{ if}\mathrm{\angle }A=\mathrm{\angle }E$$

The required two angles are $$\mathrm{\angle }A\text{ and}\mathrm{\angle }E.$$

Sample Question 2: In triangles ABC and DEF$$,\mathrm{\angle }A=\mathrm{\angle }D,\mathrm{\angle }B=\mathrm{\angle }E\text{ and }AB=EF$$. Will the two triangles be congruent? Give reasons for your answer. 

Ans: In triangles ABC and DEF$$,\mathrm{\angle }A=\mathrm{\angle }D,\mathrm{\angle }B=\mathrm{\angle }E\text{ and AB}\text{ = }\text{EF}$$.

Here, $$\text{AB and EF}$$ are not corresponding sides that touch the same two angle pairs in the two triangles.

Hence, the two triangles ABC and DEF will not be congruent.

EXERCISE 7.2

1. In triangles $$ABC$$ and $$PQR,\mathrm{\angle }A=\mathrm{\angle }Q$$ and $$\mathrm{\angle }B=\mathrm{\angle }R$$. Which side of $\mathrm{\Delta }PQR$ should be equal to side $AB$ of $\mathrm{\Delta }ABC$ so that the two triangles are congruent? Give reason for your answer.

Ans: In triangles $$\text{ABC and PQR},\mathrm{\angle }A=\mathrm{\angle }Q\text{ and}\mathrm{\angle }B=\mathrm{\angle }R$$.

The corresponding sides needs to be equal for the triangles to be congruent.

$$\therefore \text{AB}\text{ = }\text{PQ}$$

Therefore,$$\mathrm{\Delta }ABC\cong \mathrm{\Delta }PQR\left(\text{SAS axiom of congruency}\right)$$

2. In triangles $$ABC$$ and $$PQR,\mathrm{\angle }A=\mathrm{\angle }Q$$ and$$\mathrm{\angle }B=\mathrm{\angle }R.$$ Which side of $$\mathrm{\Delta }PQR$$ should be equal to side BC of $$\mathrm{\Delta }ABC$$ so that the two triangles are congruent? Give reason for your answer.

Ans: In triangles $$\text{ABC and PQR},\mathrm{\angle }A=\mathrm{\angle }Q\text{ and}\mathrm{\angle }B=\mathrm{\angle }R.$$

We know that:

The two triangles are congruent if the corresponding side that touches the same two angle pairs is equal in the given pair of triangles.

Here,$$\text{BC}\text{ = }\text{RP}$$                          (Corresponding side.)

Therefore, $$\mathrm{\Delta }ABC\cong \mathrm{\Delta }PQR\left(\text{SAS axiom of congruency}\right)$$

3. “If two sides and an angle of one triangle are equal to two sides and an angle of Another triangle, then the two triangles must be congruent.” Is the statement true? Why?

Ans: The given statement is “If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.”

We know that:

If two sides and the included angle of one triangle is equal to corresponding two sides and the included angle of another triangle then the two triangles must be congruent.

Hence, the given statement is false.

4. “If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent.” Is the statement true? Why?

Ans: The statement is “If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent.”

We know that:

If two angles and the corresponding side of one triangle are equal to the two included angles and the corresponding side of another triangle then the two triangles must be congruent. 

The given statement is false.

5. Is it possible to construct a triangle with lengths of its sides as $$4cm,3cm$$ and $$7cm$$? Give reason for your answer.

Ans: Given 3 sides are $$4\text{ cm},3\text{ cm and }7\text{ cm}$$.

To construct the triangle of given sides, we need to check property of triangle “The sum of any two sides of a triangle is always greater than the third side" is satisfied by the lengths of all the three sides.

Here, $4+3\ngtr 7$

Hence, it is not possible to construct a triangle with the given lengths $$4\text{ cm},3\text{ cm and }7\text{ cm}$$.

6. It is given that $$\mathrm{\Delta }ABC\cong \mathrm{\Delta }RPQ$$. Is it true to say that $$\text{BC}\text{ = }\text{QR}$$? Why?

Ans: It is given that $$\mathrm{\Delta }ABC\cong \mathrm{\Delta }RPQ$$.

We know that:

The two triangles are congruent, if the sides and angles of one triangle are equal to the corresponding side and angles of other triangle.
For $$\mathrm{\Delta }ABC\cong \mathrm{\Delta }RPQ$$.
We have,  $$AB=RP,BC=PQ$$ and $$AC=RQ$$, 

Hence, it is not true to say that $$BC=QR$$.

7. If $$\mathrm{\Delta }PQR\cong \mathrm{\Delta }EDF$$, then is it true to say that $$PR=EF$$? Give reason for your Answer.

Ans: $$\mathrm{\Delta }PQR\cong \mathrm{\Delta }EDF$$

$$\text{For }\mathrm{\Delta }PQR\cong \mathrm{\Delta }EDF$$, we have

 $$\text{PR}\text{ = }\text{EF}$$           (Corresponding sides)

8. In $$\mathrm{\Delta }PQR$$, $$\mathrm{\angle }P={70}^{\circ }$$ and $$\mathrm{\angle }R={30}^{\circ }$$. Which side of this triangle is the longest? Give Reason for your answer.

Ans: In $$\mathrm{\Delta }PQR$$, $$\mathrm{\angle }P={70}^{\circ },\mathrm{\angle }R={30}^{\circ }$$

By Angle sum property of triangle, we have:

$$\mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R={180}^{\circ }$$     (sum of angles in a triangle is $${180}^{\circ }$$)

$${100}^{\circ }+\mathrm{\angle }Q={180}^{\circ }$$

$$\mathrm{\angle }Q={180}^{\circ }-{100}^{\circ }={80}^{\circ }$$

$$\mathrm{\angle }Q={80}^{\circ }$$ the greatest angle.   

Its opposite side $$\text{PR}$$

  We know that, the side opposite greatest angle is longest.

The side $$\text{PR}$$ is the longest side.

9.$$AD$$ is a median of the triangle$$\text{ABC}$$. Is it true that$$AB+BC+CA>2AD$$? Give Reason for your answer.

Ans: $$\text{AD}$$ is a median of the triangle $$\text{ABC}$$.

The median AD of a triangle $$\text{ABC}$$divides it into two triangles of equal areas $$\mathrm{\Delta }ABD$$ and $$\mathrm{\Delta }ACD$$.

 In $$\mathrm{\Delta }ABD$$,

$$\text{AB}\text{ + }\text{BD}\text{ > }\text{AD}$$                          (Sum of two sides of triangle is greater than the third side)

 In $$\mathrm{\Delta }ACD$$,

$$\text{AC}\text{ + }\text{CD}\text{ > }\text{AD}$$                                    (Sum of two sides of triangle is greater than the third side)

On adding above inequalities,

$$\text{AB}\text{ + }\text{BD}\text{ + }\text{AC}\text{ + }\text{CD}\text{ > }\text{2AD}$$

$$\text{AB}\text{ + }\text{AC}\text{ + }\text{BC}\text{ > }\text{2AD}$$

Hence, the statement “$$\text{AB + BC + CA > 2 AD}$$” is true.

10. $$M$$ is a point on side $$BC$$ of a triangle $$ABC$$ such that $$AM$$ is the bisector of $$\mathrm{\angle }BAC$$. Is it true to say that perimeter of the triangle is greater than $$2AM$$? Give reason for your answer.

Ans: $$\text{M}$$ is a point on side $$\text{BC}$$ of a triangle $$\text{ABC}$$ such that $$\text{AM}$$ is the bisector of $$\mathrm{\angle }BAC$$. The triangle $$\text{ABC}$$divides it into two triangles $$\mathrm{\Delta }ABM$$ and $$\mathrm{\Delta }ACM$$.

In $$\mathrm{\Delta }ABM$$

The sum of the two sides of a triangle is always greater that the third side.

$$\text{AB}\text{ + }\text{BM}\text{ > }\text{AM}$$                             …... (1)

In $$\mathrm{\Delta }ACM$$,

The sum of the two sides of a triangle is always greater that the third side.

$$\text{AC}\text{ + }\text{CM}\text{ > }\text{AM}$$ ......(2)

Now, adding equation (1) and (2),

$$\text{AB}\text{ + }\text{BM}\text{ + }\text{AC}\text{ + }\text{CM}\text{ > }\text{AM}\text{ + }\text{AM}$$

$$\text{AB}\text{ + }\text{AC}\text{ + }\text{BC}\text{ > }\text{2AM}$$

Hence, the perimeter of $$\mathrm{\Delta }ABC$$ is greater than$$\text{2AM}$$.

11. Is it possible to construct a triangle with lengths of its sides as $$9\text{ cm},7\text{ cm and }17\text{ cm}$$? Give reason for your answer.

Ans: Given sides, $$9\text{ cm},7\text{ cm and }17\text{ cm}$$

To construct the triangle of given sides, we need to check property of triangle “The sum of any two sides of a triangle is always greater than the third side" is satisfied by the lengths of all the three sides.

Here, $$\text{9}\text{ + }\text{7}\text{ = }\text{16}\text{ < }\text{17}$$

The sum of two sides is less than the third side. Hence, it is not possible to construct a triangle with lengths of its sides as $$9\text{ cm},7\text{ cm and }17\text{ cm}$$.

12. Is it possible to construct a triangle with lengths of its sides as $$\text{8 cm},7\text{ cm and 4 cm}$$? Give reason for your answer.

Ans: Given sides, $$\text{8 cm},7\text{ cm and 4 cm}$$

To construct the triangle of given sides, we need to check property of triangle “The sum of any two sides of a triangle is always greater than the third side" is satisfied by the lengths of all the three sides.

Here, 

$$8+7=15>4$$

$$8+4=12>7$$

$$7+4=11>8$$

The sum of two sides is less than the third side. 

Hence, it is possible to construct a triangle with lengths of its sides as $$\text{8 cm},7\text{ cm and 4 cm}$$

Sample Question 1: In Fig$$,PQ=PR$$ and $$\mathrm{\angle }Q=\mathrm{\angle }R.$$

Prove that $$\mathrm{\Delta }PQS\cong \mathrm{\Delta }PRT$$

Ans: In $$\mathrm{\Delta }PQSand\mathrm{\Delta }PRT$$,

  $\text{PQ}\text{ = PR }\left(\text{Given}\right)$

  $\mathrm{\angle }Q=\mathrm{\angle }R\left(\text{Given}\right)$ 

and  $$\mathrm{\angle }QPS=\mathrm{\angle }RPT$$ (same angle) 

Therefore$$,\mathrm{\Delta }PQS\cong \mathrm{\Delta }PRT$$ (ASA criterion of congruence.)

Sample Question 2: In Fig., two lines $$\text{AB}$$ and $$\text{CD}$$ intersect each other at the point $\text{O}$ such that $\text{BC}\text{|}\text{|}\text{AD}$ and $$\text{BC = DA}$$. Show that $\text{O}$ is the midpoint of both the line-segments $$\text{AB}$$ and $$\text{CD}$$.

Ans: The two lines $$\text{AB}$$and $$\text{CD}$$intersect each other at the point $\text{O}$ such that $\text{BC}\text{|}\text{|}\text{AD}$ 

$$\text{BC||AD}$$     (Given) 

We know that:

If a transversal intersects two parallel lines, then alternate interior angles are equal and conversely.

Therefore, $$\mathrm{\angle }CBO=\mathrm{\angle }DAO$$ (Alternate interior angles)

and $$\mathrm{\angle }BCO=\mathrm{\angle }ADO$$ (Alternate interior angles)

Also, $$\text{BC = DA}$$           (Given)

 So, $$\mathrm{\Delta }BOC\cong \mathrm{\Delta }AOD$$ (ASA)

 Therefore, $$\text{OB}\text{ = }\text{OA}$$ and $$\text{OC}\text{ = }\text{OD}$$, i.e., $$\text{O}$$ is the mid-point of both $$\text{AB and CD}$$.

Sample Question 3: In Fig, $$PQ>PR$$ and QS and RS are the bisectors of $$\mathrm{\angle }Q$$  and $$\mathrm{\angle }R,$$ respectively. Show that $$SQ>SR.$$

Ans: $$\text{PQ > PR}$$            (Given) 

Therefore$$,\mathrm{\angle }R>\mathrm{\angle }Q$$           (Angles opposite to the longer side is greater) 

$${\displaystyle \frac{1}{2}}\mathrm{\angle }R>{\displaystyle \frac{1}{2}}\mathrm{\angle }Q$$           (multiply${\displaystyle \frac{1}{2}}$ on both sides)

$$\mathrm{\angle }SRQ>\mathrm{\angle }SQR$$      $$\left(\text{as SQ and SR are the angle bisectors}\right)$$

Therefore$$,SQ>SR$$            (Side opposite the greater angle will be longer)

EXERCISE 7.3

1. ABC is an isosceles triangle with AB  =  AC and BD and CE are its two medians. Show that $$BD=CE$$

Ans: Given:

$$\text{AB = AC}$$

Also, $$\text{BD and CE}$$ are two medians.

Hence, $\text{E}$ is the midpoint of $$\text{AB}$$ and $\text{D}$ is the midpoint of $$\text{CE}$$

$${\displaystyle \frac{1}{2}}AB={\displaystyle \frac{1}{2}}AC$$

$$\text{BE = CD}$$

  In $$\mathrm{\Delta }BEC$$ and $$\mathrm{\Delta }CDB$$,

$$\text{BE = CD}$$                   (Given)

$$\mathrm{\angle }EBC=\mathrm{\angle }DCB$$                               (Angles opposite to equal sides $$\text{AB}$$and$$\text{AC}$$)

$$\text{BC = CB}$$                                     ( Common )

By SAS axiom, we have:

 $$\mathrm{\Delta }BEC\cong \mathrm{\Delta }CDB$$                               

 $$\text{BD = CE}$$                                         (by cpct)

2. In Fig.7.4, D and E are points on side $$BC$$ of a $$\mathrm{\Delta }ABC$$ such that BD  =  CE and AD  =  AE. Show that $$\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACE$$.

Ans: According to the question,

In$$\mathrm{\Delta }ABC$$,

$$\text{BD = CE}\text{and AD = AE}$$.

In$$\mathrm{\Delta }ADE$$,

$$\text{AD = AE}$$

Since opposite angles to equal sides are equal,

We have,

$$\mathrm{\angle }ADE=\mathrm{\angle }AED...\left(1\right)$$

Now, $$\mathrm{\angle }ADE+\mathrm{\angle }ADB={180}^{\circ }$$(linear pair)

$$\mathrm{\angle }ADB={180}^{\circ }-\mathrm{\angle }ADE$$$$...\left(2\right)$$

Also, $$\mathrm{\angle }AED+\mathrm{\angle }AEC={180}^{\circ }$$(linear pair)

$$\mathrm{\angle }AEC=\text{1}{80}^{\circ }-\mathrm{\angle }AED$$

$$\text{Since},\mathrm{\angle }ADE=\mathrm{\angle }AED$$

$$\mathrm{\angle }AEC={180}^{\circ }-\mathrm{\angle }ADE...\left(3\right)$$

From equation (2) and (3)

$$\mathrm{\angle }ADB=\mathrm{\angle }AEC...\left(4\right)$$

Now, In  $$\mathrm{\Delta }ADB$$ and $$\mathrm{\Delta }AEC$$,

$$\text{AD = AE}$$      (given)

$$\text{BD = EC}$$       (given)

$$\mathrm{\angle }ADB=\mathrm{\angle }AEC$$( from (4))

Hence, $$\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACE$$(by SAS)

3. $$CDE$$ is an equilateral triangle formed on a side $$CD$$ of a square $$ABCD$$ (Fig.7.5). Show that $$\mathrm{\Delta }ADE\cong \mathrm{\Delta }BCE$$.

Ans: Given $$\text{ABCD}$$ is a square

So $$\text{AB = BC = CD = AD}$$

Now in $$\mathrm{\Delta }EDC$$ is an equilateral triangle.

So $$\text{DE = EC = CB}$$

In $$\mathrm{\Delta }AED$$ and $$\mathrm{\Delta }CEB$$

$$\text{AD = BC}$$          (Side of triangle)

$$\text{DE = CE}$$           (Side of equilateral triangle)

$$\mathrm{\angle }ADE=\mathrm{\angle }ADC+\mathrm{\angle }CDE$$

  $={90}^{\circ }+{60}^{\circ }$

  $={150}^{\circ }$ 

And,

$$\mathrm{\angle }BCE=\mathrm{\angle }BCD+\mathrm{\angle }DCE$$

  $={90}^{\circ }+{60}^{\circ }$ 

  $={150}^{\circ }$ 

$$\Rightarrow \mathrm{\angle }ACE=\mathrm{\angle }BCDE$$

Hence from SAS criterion of congruence $$\mathrm{\Delta }ADE\cong \mathrm{\Delta }BCE$$.

4. In Fig.7.6, $$BA\mathrm{\perp }AC,DE\mathrm{\perp }DF$$ such that BA  =  DE and BF  =  EC. Show that $$\mathrm{\Delta }ABC\cong \mathrm{\Delta }DEF.$$

Ans: According to the question,

$$BA\mathrm{\perp }AC,DE\mathrm{\perp }DF$$ such that $$\text{BA = DE and }\text{BF = EC}$$.

In $$\mathrm{\Delta }ABC$$ and $$\mathrm{\Delta }DEF$$

$$\text{BA = DE}$$ (given)

$$\text{BF = EC}$$ (given)

$$\mathrm{\angle }A=\mathrm{\angle }D\left(\text{both }{90}^{\circ }\right)$$

$$BC=BF+FC$$

$$EF=EC+FC=BF+FC\left(\because EC\text{ = }BF\right)$$

$$\Rightarrow EF=BC$$

By SSS axiom, we have:

$$\mathrm{\Delta }ABC\cong \mathrm{\Delta }DEF$$

5.$$Q$$ is a point on the side $$SR$$ of a $$\mathrm{\Delta }PSR$$ such that $$PQ=PR$$. Prove that $$PS>PQ$$.

Ans: Given: $\text{Q}$ is a point on side $$\text{SR in Delta PSR}$$such that $$\text{PQ = PR}$$

$$\Rightarrow PS>PR\dots \dots \left(1\right)$$

In $$\mathrm{\Delta }PQR$$,

$$PQ=PR$$

$$\Rightarrow \mathrm{\angle }PQR=\mathrm{\angle }PRQ$$ (angle opposite to equal sides are equal) 

 Now, from $$\left(1\right)$$

$$\mathrm{\angle }PRQ>\mathrm{\angle }PSQ\text{ or }\mathrm{\angle }PRS>\mathrm{\angle }PSR$$

Now, in $$\mathrm{\Delta }PSR$$,

$$\mathrm{\angle }PRS>\mathrm{\angle }PSR$$

$$\Rightarrow PS>PR$$                 (Side opposite greater angle is greater)

$$\Rightarrow PS>PQ$$

Hence, proved.

6. $S$ is any point on the side $$QR$$ of a $\mathrm{\Delta }PQR$. Show that: $$PQ+QR+RP>2PS$$

Ans: $\text{S}$ is any point on the side $$\text{QR}$$of a $\mathrm{\Delta }PQR$.

Join the points $$\text{P}$$ and $$\text{S}$$.

We know that:

The sum of the length of any two sides is greater than the third side.

In $$\mathrm{\Delta }PQS$$, $$PQ+QS>PS...........\left(1\right)$$

In $$\mathrm{\Delta }PSR$$,$$PR+SR>PS...........\left(2\right)$$

$$\text{on adding equation }\left(1\right)\text{ and }\left(2\right)$$,

$$PQ+QS+SR+PR>PS+PS$$

$$PQ+\left(QS+SR\right)+PR>2PS$$

$$PQ+QR+PR>2PS$$

7. D is any point on the side $$AC$$ of a $$\mathrm{\Delta }ABC$$ with AB  =  AC. Show that $$CD<BD$$.

Ans: D is any point on side $$\text{AC}$$ of a $$\mathrm{\Delta }ABC$$ with AB  =  AC.

We know that:

Opposite angles to equal sides are equal.

$$\mathrm{\angle }ABC=\mathrm{\angle }ACB......\left(1\right)$$

From the figure, we have: 

$$\mathrm{\angle }ABC>\mathrm{\angle }DBC$$

From equation $$(1)$$, we have:

$$\mathrm{\angle }ACB>\mathrm{\angle }DBC$$

i.e.$$\mathrm{\angle }DCB>\mathrm{\angle }DBC$$

It means that

$$BD>CD$$………………{Opposite side}

So we get

$$CD<BD$$

Therefore, it is proved that $$CD<BD$$.

8. In Fig. 7.7, $$l||m$$ and M is the midpoint of a line segment $$AB$$. Show that $M$ is also the mid-point of any line segment $$CD$$, having its end points on $l$ and$m$, respectively.

Ans: $$\text{AM = MB}$$                 (Given) 

$$\mathrm{\angle }DMB=\mathrm{\angle }AMC$$             (Vertically Opposite Angle)

$$\mathrm{\angle }ACD=\mathrm{\angle }CDB$$                (Alternating Angle)

Hence, According to ASA Congruence Criterion

$$\mathrm{\Delta }AMC\cong \mathrm{\Delta }DMB$$

Hence, $$\text{DM = MC}$$

So, Point $\text{M}$ is mid-point of $\text{CD}$.

9. Bisectors of the angles $B$ and $C$ of an isosceles triangle with $$AB=AC$$ intersect each other at $$O$$. $$BO$$ is produced to a point $$M$$. Prove that $$\mathrm{\angle }MOC=\mathrm{\angle }ABC$$

Ans: Given,

Lines $$\text{OB}$$ and $$\text{OC}$$ are the bisectors of $$\mathrm{\angle }B\text{ and }\mathrm{\angle }C$$ of an isosceles $$\mathrm{\Delta }ABC$$ such that $$\text{AB = AC}$$ which intersect each other at $$\text{O}$$and $\text{BO}$ is produced to $\text{M}$.

In $$\mathrm{\Delta }ABC$$,

$$\text{AB = AC}$$(given)

$$\mathrm{\angle }ACB=\mathrm{\angle }ABC$$……. The Opposite angles to equal sides are equal.

$${\displaystyle \frac{1}{2}}\mathrm{\angle }ACB={\displaystyle \frac{1}{2}}\mathrm{\angle }ABC$$           (Dividing both sides by 2)

Now, OB and OC are the bisector of $$\mathrm{\angle }B$$ and $$\mathrm{\angle }C$$

We know that:

The bisector divides a given angle into two angles with equal measures.

$$\mathrm{\angle }OCB=\mathrm{\angle }OBC\dots \dots \left(1\right)$$

Now, from equation (1), we have

$$\Rightarrow \mathrm{\angle }MOC=\mathrm{\angle }OBC+\mathrm{\angle }OBC$$

$$\Rightarrow \mathrm{\angle }MOC=\mathrm{\angle }OBC$$

$$\Rightarrow \mathrm{\angle }MOC=2\mathrm{\angle }ABC$$

(Since, $$\text{OB}$$ is the bisector of $$\mathrm{\angle }B$$)

10. Bisectors of the angles $$B$$ and $$C$$ of an isosceles triangle$$ABC$$ with $$AB=AC$$ intersect each other at $O$. Show that the external angle adjacent to $$\mathrm{\angle }ABC$$ is equal to $$\mathrm{\angle }BOC$$.

Ans: Bisectors of the angles $$\text{B}$$ and $$\text{C}$$ of an isosceles triangle$$\text{ABC}$$ with $$\text{AB = AC}$$ intersect each other at $\text{O}$

In $$\mathrm{\Delta }ABC$$

$$\text{AB = AC}$$

$$\mathrm{\angle }ACB=\mathrm{\angle }ABC$$

$${\displaystyle \frac{1}{2}}\mathrm{\angle }ACB={\displaystyle \frac{1}{2}}\mathrm{\angle }ABC$$

$$\mathrm{\angle }OCB=\mathrm{\angle }OBC$$               …. (1)

Since, BO and CO are the bisectors of $\mathrm{\angle }ABC\text{ and }\mathrm{\angle }\text{ACB}$.

In $$\mathrm{\Delta }BOC$$,

$$\mathrm{\angle }OBC+\mathrm{\angle }OCB+\mathrm{\angle }BDC={180}^{\circ }$$    (Sum of angles in a triangle is ${180}^{\circ }$)

$$2\mathrm{\angle }OBC+\mathrm{\angle }BOC={180}^{\circ }$$               from equation (1)

$$\mathrm{\angle }ABC+\mathrm{\angle }BOC={180}^{\circ }$$                 ($\because BO$ is the bisectors of $\mathrm{\angle }ABC$)

$${180}^{\circ }-\mathrm{\angle }DBA+\mathrm{\angle }BOC={180}^{\circ }$$       ($\because DBC$ is a straight line)

   $-\mathrm{\angle }DBA+\mathrm{\angle }BOC=0$

  $\mathrm{\angle }BOC=\mathrm{\angle }DBA$ 

11. In Fig, AD is the bisector of $$\mathrm{\angle }BAC$$. Prove that $$AB>BD$$

Ans: In  $$\mathrm{\Delta }ABC$$

Let $$\mathrm{\angle }1=\mathrm{\angle }BAD,\mathrm{\angle }2=\mathrm{\angle }DAC,\mathrm{\angle }3=BDA$$

AD is the bisector of $$\mathrm{\angle }BAC$$

$$\mathrm{\angle }1=\mathrm{\angle }2$$…(i)

Also in $$\mathrm{\Delta }ADC$$

$$\mathrm{\angle }3=\mathrm{\angle }2+\mathrm{\angle }C$$            (ext. angle is equal to sum of opposite interior angles)

$$\mathrm{\angle }3>\mathrm{\angle }2$$            (ext. angle is greater than one of the interior angles)

But $$\mathrm{\angle }1=\mathrm{\angle }2$$

$$\mathrm{\angle }3>\mathrm{\angle }1$$             (Sides opposite greater angle is greater)

$$AB>BD$$

Hence proved.

Sample Question 1: In Fig. 7.9, ABC is a right triangle and right angled at B such that $$\mathrm{\angle }BCA=2\mathrm{\angle }BAC$$. Show that hypotenuse $$AC=2BC$$.

Ans:  Produce $$\text{CB}$$ to a point $$\text{D}$$ such that $$\text{BC = BD}$$  and join $$\text{AD}$$.

 In $$\mathrm{\Delta }ABC$$ and $$\mathrm{\Delta }ABD$$, we have $$\text{BC = BD}$$  (By construction)

 $$\text{AB}$$ is common in both triangles

$$\mathrm{\angle }ABC=\mathrm{\angle }ABD$$ (Each of 90°)

 Therefore$$,\mathrm{\Delta }ABC\cong \mathrm{\Delta }ABD$$ (SAS)

 So, $$\mathrm{\angle }CAB=\mathrm{\angle }DAB$$                                     ..….(1) 

Let $$x=\mathrm{\angle }CAB=\mathrm{\angle }DAB$$

and  $$\text{AC = AD}$$                                             ……(2)

 Thus, 

  $\mathrm{\angle }CAD=\mathrm{\angle }CAB+\mathrm{\angle }BAD$

  $\Rightarrow x+x=2x$                  {From (1)}    …..(3)

 and $$\mathrm{\angle }ACD=\mathrm{\angle }ADB=x$$         {From (2), $$\text{AC = AD}$$}    …..(4)

 That is, $$\mathrm{\Delta }ACD$$ is an equilateral triangle.             {From (3) and (4)} 

or   $$\text{AC = CD}$$, i.e., $$\text{AC = 2BC}$$      {Since $$\text{BC = BD}$$}

Sample Question 2: Prove that if in two triangles two angles and the included side of

one triangle is equal to two angles and the included side of the other triangle, then the two triangles are congruent.

Ans: In two triangles two angles and the included side of one triangle is equal to two angles and the included side of the other triangle.

Draw the triangles.

Let say,

$$\mathrm{\angle }B=\mathrm{\angle }Q$$

$$\mathrm{\angle }C=\mathrm{\angle }R$$

$$AB=PQ$$

$$\therefore \mathrm{\Delta }ABC\cong \mathrm{\Delta }PQR$$(BY AAS)

Sample Question 3: If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles.

Ans: The bisector of an angle of a triangle also bisects the opposite side.

$\mathrm{\Delta }ABC$, $$AD$$ bisects $$\mathrm{\angle }A$$ and $\text{AD}$ bisects $\text{BC}$. 

In $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACD$,

$\mathrm{\angle }DAB=\mathrm{\angle }DAC$            ($$\text{AD}$$ bisects $\mathrm{\angle }\text{A}$)

$AD=AD(\text{Common})$

$\text{BD}=\text{CD}$    ($\text{AD}$ bisects $\text{BC})$

$\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$ (SAS rule)

Thus, $\text{AB = AC}$ (By cpct) 

We know that:

A triangle that has two sides of equal length is known as Isosceles triangle.

Hence, $\mathrm{\Delta }ABD$ is an Isosceles triangle.

Sample Question 4: $$\text{S}$$ is any point in the interior of $$\mathrm{\Delta }PQR$$. Show that $$SQ+SR<PQ+PR.$$

Ans: $$\text{S}$$ is any point in the interior of $$\mathrm{\Delta }PQR$$.

Let us project $\text{QS}$ to intersect $\text{PR}$ at $\text{T}$.

 From $\mathrm{\Delta }PQT$, we have $PQ+PT>QT$

$PQ+PT>SQ+ST$ …(i)

$ST+TR>SR$ …(ii)

Adding equation (1) and (2), 

$PQ+PT+ST+TR>SQ+ST+SR$

$PQ+PT+TR>SQ+SR$

$PQ+PR>SQ+SR$

$$\Rightarrow SQ+SR<PQ+PR.$$

EXERCISE 7.4

1. Find all the angles of an equilateral triangle.

Ans: Given,

$$\mathrm{\Delta }ABC$$ is an equilateral triangle. 

In an equilateral triangle all the angles are equal.

$$\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C=x$$

Sum of angles in a triangle $$={180}^{\circ }$$

$$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$$

$$x+x+x={180}^{\circ }$$

$$x={60}^{\circ }$$

Thus, $$\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C={60}^{\circ }$$

2. The image of an object placed at a point $$A$$ before a plane mirror $$LM$$ is seen at the point $B$ by an observer at $$D$$ as shown in Fig. Prove that the image is as far behind the mirror as the object is in front of the mirror. 

Ans: The image of an object placed at a point $$\text{A}$$ before a plane mirror $$\text{LM}$$ is seen at the point $\text{B}$ by an observer at $$\text{D}$$ as shown in Fig.

From figure we need to prove that AT=BT

We have,

Angle of incidence = Angle of reflection

Therefore,

$$\mathrm{\angle }ACN=\mathrm{\angle }DCN.....\left(1\right)$$

Since, $$\text{AB}\parallel \text{CN}$$ and $$\text{AC}$$ is the transversal.

From the figure we know that $$\mathrm{\angle }TAC\text{ and }\mathrm{\angle }ACN$$ are alternate angles.

$$\mathrm{\angle }TAC=\mathrm{\angle }CAN.....\left(2\right)$$

We know that $$\text{AB}\parallel \text{CN and BD}$$ is the transversal.

From the figure we know that $$\mathrm{\angle }TBC\text{ and }\mathrm{\angle }DCN$$ are corresponding angles.

$$\mathrm{\angle }TBC=\mathrm{\angle }DCN.....\left(3\right)$$

By considering the equation $$\left(1\right),\left(2\right)\text{ and }\left(3\right)$$

We get,

$$\mathrm{\angle }TAC=\mathrm{\angle }TBC.....(4)$$

Now in $$\mathrm{\Delta }ACT$$ and $$\mathrm{\Delta }BCT$$

$$\mathrm{\angle }ATC=\mathrm{\angle }BTC={90}^{\circ }$$

$\text{CT}$ is common i.e., $$\text{CT = CT}$$

By AAS congruence criterion

$$\mathrm{\Delta }ACT\cong \mathrm{\Delta }BCT$$

$$\text{AT = BT}\left(\text{c}\text{.p}\text{.c}\text{.t}\right)$$

Therefore, it is proved that the image is as far behind the mirror as the object is in front of the mirror.

3. $$ABC$$ is an isosceles triangle with $$AB=AC$$ and $$D$$ is a point on $$BC$$ such that $$AD\mathrm{\perp }BC$$ (Fig. 7.13). To prove that $$\mathrm{\angle }BAD=\mathrm{\angle }CAD$$, a student proceeded as follows:

In $$\mathrm{\Delta }ABD$$ and $$\mathrm{\Delta }ACD$$,

$$AB=AC$$ (Given) 

$$\mathrm{\angle }B=\mathrm{\angle }C$$ (because$$AB=AC$$)

 and $$\mathrm{\angle }ADB=\mathrm{\angle }ADC$$

Therefore, $$\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$$ (AAS)

 So, $$\mathrm{\angle }BAD=\mathrm{\angle }CAD$$ (cpct) 

What is the defect in the above arguments? 

{Hint: Recall how $$\mathrm{\angle }B=\mathrm{\angle }C$$ is proved when $$AB=AC$$}.

Ans: $\text{In }\mathrm{\Delta }ABC,$

  $AB=AC$

  $\Rightarrow \mathrm{\angle }ACB=\mathrm{\angle }ABC$

  $\text{In }\mathrm{\Delta }ABD\text{ and }\mathrm{\Delta }ACD,$ 

   $AB=AC$     (given)

   $\mathrm{\angle }ABD=\mathrm{\angle }ACD$      (Proved above)

   $\mathrm{\angle }ADB=\mathrm{\angle }ADC$      (each ${90}^{\circ }$)

$\therefore \mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$       (by AAS)

So, $\mathrm{\angle }BAD=\mathrm{\angle }CAD$        (by CPCT)

So, the defect in the given argument is that firstly prove $\mathrm{\angle }ABD=\mathrm{\angle }ACD$ Hence, $\mathrm{\angle }ABD=\mathrm{\angle }ACD$ is a defect.

4. P is a point on the bisector of $$\mathrm{\angle }ABC$$. If the line through$$P$$, parallel to $$BA$$ meet $$BC$$at $$Q$$, prove that $$BPQ$$ is an isosceles triangle.

Ans: Let BP is bisector of $\mathrm{\angle }ABC$

$\Rightarrow \mathrm{\angle }ABP=\mathrm{\angle }PBC$ …….(1)

Now,  (given) 

$\Rightarrow \mathrm{\angle }BPQ=\mathrm{\angle }ABP\dots .(2)$  {Alternative angles}

From (1) and (2), we get

$\mathrm{\angle }BPQ=\mathrm{\angle }PBC$ Or $\mathrm{\angle }BPQ=\mathrm{\angle }PBQ$

Now, In $\mathrm{\Delta }BPQ$

$\mathrm{\angle }BPQ=\mathrm{\angle }PBQ$

$$\Rightarrow \mathrm{\Delta }\text{BPQ}$$ is an isosceles triangle.

Hence proved.

5. ABCD is a quadrilateral in which AB  =  BC and AD  =  CD. Show that $$BD$$ bisects both the angles ABC and ADC.

Ans: ABCD is a quadrilateral in which $$\text{AB = BC and AD = CD}\text{.}$$

In$$\mathrm{\Delta }ABD$$ and $$\mathrm{\Delta }CBD$$,

$$\text{AD = CD}$$       (given )