# RD Sharma Solutions for Class 9 Maths Chapter 10 Congruent Triangles

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Chapter 10 of RD Sharma Class 9 deals with Congruent Triangles, their different types and formulas. If you want to improve your basic in Congruent Triangles, then you can use this.

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## RD Sharma Solutions for Class 9 Maths Chapter 10 Congruent Triangles - Free PDF Download

Question 1: In figure, the sides BA and CA have been produced such that $B A=A D$ and $C A=$ AE. Prove that segment DE ॥ BC.

Solution: Sides $B A$ and $C A$ have been produced such that $B A=A D$ and $C A=A E$.

To prove: DE ॥ BC

Consider $\triangle \mathrm{BAC}$ and $\triangle \mathrm{DAE}$

$\mathrm{BA}=\mathrm{AD}$ and $\mathrm{CA}=\mathrm{AE}$ (Given)

$\angle B A C=\angle D A E$ (vertically opposite angles)

By SAS congruence criterion, we have

$\triangle \mathrm{BAC} \simeq \triangle \mathrm{DAE}$

We know, corresponding parts of congruent triangles are equal

So, $\mathrm{BC}=\mathrm{DE}$ and $\angle \mathrm{DEA}=\angle \mathrm{BCA}, \angle \mathrm{EDA}=\angle \mathrm{CBA}$

Now, DE and BC are two lines intersected by a transversal DB s.t.

$\angle \mathrm{DEA}=\angle \mathrm{BCA}$ (alternate angles are equal)

Therefore, DE ॥ BC. Proved.

Question 2: In a PQR, if $P Q=Q R$ and $L, M$ and $N$ are the mid-points of the sides $P Q, Q R$ and RP respectively. Prove that $L N=M N$.

Solution: Draw a figure based on given instruction,

In $\triangle \mathrm{PQR}, \mathrm{PQ}=\mathrm{QR}$ and $\mathrm{L}, \mathrm{M}, \mathrm{N}$ are midpoints of the sides $\mathrm{PQ}, \mathrm{QP}$ and $\mathrm{RP}$ respectively (Given)

To prove : $L N=M N$

As two sides of the triangle are equal, so $\triangle \mathrm{PQR}$ is an isosceles triangle

$\mathrm{PQ}=\mathrm{QR}$ and $\angle \mathrm{QPR}=\angle \mathrm{QRP} \ldots \ldots$ (i)

Also, $L$ and $M$ are midpoints of $P Q$ and $Q R$ respectively

$\mathrm{PL}=\mathrm{LQ}=\mathrm{QM}=\mathrm{MR}=\mathrm{QR} / 2$

Now, consider $\triangle$ LPN and $\Delta$ MRN,

$\mathrm{LP}=\mathrm{MR}$

$\angle \mathrm{LPN}=\angle \mathrm{MRN}$ [From (i)]

$\angle \mathrm{QPR}=\angle \mathrm{LPN}$ and $\angle \mathrm{QRP}=\angle \mathrm{MRN}$

$\mathrm{PN}=\mathrm{NR}[\mathrm{N}$ is midpoint of $\mathrm{PR}]$

By SAS congruence criterion,

$\Delta \mathrm{LPN} \simeq \triangle \mathrm{MRN}$

We know, corresponding parts of congruent triangles are equal.

So $L N=M N$

Proved.

Question 3: In figure, PQRS is a square and SRT is an equilateral triangle. Prove that

(i) $P T=Q T$ (ii) $\angle T Q R=15^{\circ}$

Solution: Given: PQRS is a square and SRT is an equilateral triangle.

To prove:

(i) $\mathrm{PT}=\mathrm{QT}$ and (ii) $\angle \mathrm{TQR}=15^{\circ}$

Now,

PQRS is a square:

$\mathrm{PQ}=\mathrm{QR}=\mathrm{RS}=\mathrm{SP} \ldots$(i)

And $\angle S P Q=\angle P Q R=\angle Q R S=\angle R S P=90^{\circ}$

Also, $\triangle$ SRT is an equilateral triangle:

$\mathrm{SR}=\mathrm{R} \mathrm{T}=\mathrm{TS}$.$\ldots \ldots$ (ii)

And $\angle T S R=\angle S R T=\angle R T S=60^{\circ}$

From (i) and (ii)

$\mathrm{PQ}=\mathrm{QR}=\mathrm{SP}=\mathrm{SR}=\mathrm{RT}=\mathrm{TS} \ldots \ldots$(iii)

From figure,

$\angle T S P=\angle T S R+\angle R S P$

$=60^{\circ}+90^{\circ}=150^{\circ}$ and

$\angle \mathrm{TRQ}=\angle \mathrm{TRS}+\angle \mathrm{SRQ}$

$=60^{\circ}+90^{\circ}=150^{\circ}$

$\Rightarrow \angle \mathrm{TSP}=\angle \mathrm{TRQ}=150^{\circ} \ldots \ldots \ldots \ldots \ldots \ldots .$ (iv)

By SAS congruence criterion, $\triangle$ TSP $\simeq \triangle$ TRQ

We know, corresponding parts of congruent triangles are equal

So, $\mathrm{PT}=\mathrm{QT}$

Proved part (i).

Now, consider $\Delta$ TQR.

$\mathrm{QR}=$ TR [From (iii)]

$\triangle \mathrm{TQR}$ is an isosceles triangle.

$\angle \mathrm{QTR}=\angle \mathrm{TQR}$ [angles opposite to equal sides $]$

Sum of angles in a triangle $=180$ 。

$\Rightarrow \angle Q T R+\angle T Q R+\angle T R Q=180^{\circ}$

$\Rightarrow 2 \angle \mathrm{TQR}+150^{\circ}=180^{\circ}[\mathrm{From}(\mathrm{iv})]$

$=>2 \angle \mathrm{TQR}=30^{\circ}$

$=>\angle \mathrm{TQR}=15^{\circ}$

Hence proved part (ii).

Question 4: Prove that the medians of an equilateral triangle are equal.

Solution: Consider an equilateral $\triangle \mathrm{ABC}$, and Let $\mathrm{D}, \mathrm{E}, \mathrm{F}$ are midpoints of $\mathrm{BC}, \mathrm{CA}$ and $\mathrm{AB}$.

Here, $A D, B E$ and $C F$ are medians of $\triangle A B C$.

Now,

$D$ is midpoint of $B C=>B D=D C$

Similarly. $C E=E A$ and $A F=F B$

Since $\triangle \mathrm{ABC}$ is an equilateral triangle

$\mathrm{AB}=\mathrm{BC}=\mathrm{CA} \ldots$. (i)

$B D=D C=C E=E A=A F=F B \ldots \ldots \ldots$.(ii)

And also, $\angle \mathrm{ABC}=\angle \mathrm{BCA}=\angle \mathrm{CAB}=60^{\circ} \ldots \ldots \ldots$(iii)

Consider $\triangle \mathrm{ABD}$ and $\triangle \mathrm{BCE}$

$A B=B C$ [From (i)]

$\mathrm{BD}=\mathrm{CE}$ [From (ii) $]$

$\angle \mathrm{ABD}=\angle \mathrm{BCE}$ [From (iii)]

By SAS congruence criterion,

$\triangle \mathrm{ABD} \simeq \triangle \mathrm{BCE}$

$\Rightarrow A D=B E \ldots \ldots$(iv)

[Corresponding parts of congruent triangles are equal in measure]

Now, consider $\triangle \mathrm{BCE}$ and $\triangle \mathrm{CAF}$,

$\mathrm{BC}=\mathrm{CA}$ [From (i)]

$\angle B C E=\angle$ CAF [From (iii)]

$\mathrm{CE}=\mathrm{AF}[\mathrm{From}$ (ii) $]$’

By SAS congruence criterion,

$\triangle \mathrm{BCE} \simeq \triangle \mathrm{CAF}$

$=>\mathrm{BE}=\mathrm{CF} \ldots \ldots$(v)

[Corresponding parts of congruent triangles are equal] From (iv) and (v), we have

$A D=B E=C F$

Median $\mathrm{AD}=$ Median $\mathrm{BE}=$ Median $\mathrm{CF}$

The medians of an equilateral triangle are equal.

Hence proved

Question 5: In a $\triangle \mathrm{ABC}$, if $\angle \mathrm{A}=120^{\circ}$ and $\mathrm{AB}=\mathrm{AC}$. Find $\angle \mathrm{B}$ and $\angle \mathrm{C}$.

Solution:

To find: $\angle B$ and $\angle C$.

Here, $\triangle \mathrm{ABC}$ is an isosceles triangle since $\mathrm{AB}=\mathrm{AC}$

$\angle B=\angle C \ldots \ldots \ldots$…… (i)

[Angles opposite to equal sides are equal]

We know, sum of angles in a triangle $=180^{\circ}$

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

$\angle A+\angle B+\angle B=180^{\circ}$ (using (i)

$120^{\circ}+2 \angle B=180^{\circ}$

$2 \angle B=180^{\circ}-120^{\circ}=60^{\circ}$

$\angle B=30^{\circ}$

Therefore, $\angle B=\angle C=30^{\circ}$

Question 6: In a $\triangle A B C$, if $A B=A C$ and $\angle B=70^{\circ}$, find $\angle A$.

Solution: Given: In a $\triangle A B C, A B=A C$ and $\angle B=70^{\circ}$

$\angle B=\angle C$ [Angles opposite to equal sides are equal]

Therefore, $\angle B=\angle C=70^{\circ}$

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

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

$\angle A+70^{\circ}+70^{\circ}=180^{\circ}$

$\angle A=180^{\circ}-140^{\circ}$

$\angle A=40^{\circ}$

Exercise $10.2$

Question 1: In figure, it is given that $\mathrm{RT}=\mathrm{TS}, \angle 1=2 \angle 2$ and $\angle 4=2(\angle 3)$. Prove that $\Delta \mathrm{RBT} \cong$ \DeltaSAT.

Solution: In the figure,

$\mathrm{RT}=\mathrm{TS} \ldots \ldots(\mathrm{i})$

$\angle 1=2 \angle 2 \ldots \ldots$(ii)

And $\angle 4=2<3 \ldots \ldots$(iii)

To prove: $\Delta \mathrm{RBT} \cong \Delta \mathrm{SAT}$

Let the point of intersection $\mathrm{RB}$ and SA be denoted by O

$\angle \mathrm{AOR}=\angle \mathrm{BOS}$ [Vertically opposite angles $]$

or $\angle 1=\angle 4$

$2 \angle 2=2 \angle 3$ [From (ii) and (iii)]

or $\angle 2=\angle 3 \ldots \ldots$(iv)

Now in $\Delta$ TRS, we have $\mathrm{RT}=\mathrm{TS}$

$\Rightarrow \Delta \mathrm{TRS}$ is an isosceles triangle

$\angle \mathrm{TRS}=\angle \mathrm{TSR} \ldots \ldots(\mathrm{v})$

But, $\angle$ TRS $=\angle T R B+\angle 2 \ldots \ldots$ (vi)

$\angle \mathrm{TSR}=\angle \mathrm{TSA}+\angle 3 \ldots \ldots(\mathrm{vii})$

Putting (vi) and (vii) in (v) we get

$\angle T R B+\angle 2=\angle T S A+\angle 3$

$\Rightarrow \angle$ TRB $=\angle$ TSA [From (iv)]

Consider $\triangle$ RBT and $\triangle$ SAT

$\mathrm{RT}=\mathrm{ST}[$ From (i) $]$

$\angle \mathrm{TRB}=\angle \mathrm{TSA}[\mathrm{From}(\mathrm{iv})]$

By ASA criterion of congruence, we have

$\Delta \mathrm{RBT} \cong \Delta \mathrm{SAT}$

Question 2: Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines $\mathrm{AB}$ and $\mathrm{CD}$ bisect at $\mathrm{O} .$

Solution: Lines $A B$ and $C D$ Intersect at $O$

Such that BC $\|$ AD and

$\mathrm{BC}=\mathrm{AD} \ldots \ldots \text {.(i) }$

To prove : $A B$ and $C D$ bisect at $O$.

First we have to prove that $\triangle \mathrm{AOD} \cong \Delta \mathrm{BOC}$

$\angle \mathrm{OCB}=\angle \mathrm{ODA}[\mathrm{AD} \| \mathrm{BC}$ and $\mathrm{CD}$ is transversal]

$\mathrm{AD}=\mathrm{BC}[$ from $(\mathrm{i})]$

$\angle \mathrm{OBC}=\angle \mathrm{OAD}[\mathrm{AD} \| \mathrm{BC}$ and $\mathrm{AB}$ is transversal]

By ASA Criterion:

$\triangle \mathrm{AOD} \cong \Delta \mathrm{BOC}$

$\mathrm{OA}=\mathrm{OB}$ and $\mathrm{OD}=\mathrm{OC}$ (By c.p.c.t.)

Therefore, $A B$ and $C D$ bisect each other at $O$.

Hence Proved.

Question 3: BD and CE are bisectors of $\angle B$ and $\angle C$ of an isosceles $\triangle A B C$ with $A B=A C$. Prove that $B D=C E$.

Solution: $\triangle A B C$ is isosceles with $A B=A C$ and $B D$ and $C E$ are bisectors of $\angle B$ and $\angle C$ We have to prove $B D$ $=\mathrm{CE}$. (Given)

Since $\mathrm{AB}=\mathrm{AC}$

$\Rightarrow \angle \mathrm{ABC}=\angle \mathrm{ACB} \ldots \ldots(\mathrm{i})$

[Angles opposite to equal sides are equal]

Since $\mathrm{BD}$ and $\mathrm{CE}$ are bisectors of $\angle \mathrm{B}$ and $\angle \mathrm{C}$

$\angle A B D=\angle D B C=\angle B C E=E C A=\angle B / 2=\angle C / 2$...(ii)

Now, Consider $\Delta \mathrm{EBC}=\Delta \mathrm{DCB}$

$\angle E B C=\angle D C B$ [From (i)]

$\mathrm{BC}=\mathrm{BC}$ [Common side]

$\angle \mathrm{BCE}=\angle \mathrm{CBD}$ [From (ii)]

By ASA congruence criterion, $\triangle \mathrm{EBC} \cong \triangle \mathrm{DCB}$

Since corresponding parts of congruent triangles are equal.

$=>\mathrm{CE}=\mathrm{BD}$

or, $B D=C E$

Hence proved.

Exercise $10.3$

Question 1: In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.

Solution: In two right triangles one side and acute angle of one are equal to the corresponding side and angles of the other. (Given)

To prove: Both the triangles are congruent.

Consider two right triangles such that

$\angle B=\angle E=90^{\circ} \ldots \ldots$(i)

$\mathrm{AB}=\mathrm{DE} \ldots$(ii)

$\angle C=\angle F \ldots \ldots$ ……(iii)

Here we have two right triangles, $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$

From (i), (ii) and (iii),

By AAS congruence criterion, we have $\triangle \mathrm{ABC} \cong \Delta \mathrm{DEF}$

Both the triangles are congruent. Hence proved.

Question 2: If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.

Solution: Let $\mathrm{ABC}$ be a triangle such that $\mathrm{AD}$ is the angular bisector of exterior vertical angle, $\angle \mathrm{EAC}$ and $\mathrm{AD}$ ॥ BC.

From figure,

$\angle 1=\angle 2[\mathrm{AD}$ is a bisector of $\angle \mathrm{EAC}]$

$\angle 1=\angle 3$ [Corresponding angles]

and $\angle 2=\angle 4$ [alternative angle]

From above, we have $\angle 3=\angle 4$

This implies, $A B=A C$

Two sides $A B$ and $A C$ are equal.

$=>\triangle \mathrm{ABC}$ is an isosceles triangle.

Question 3: In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

Solution: Let $\triangle \mathrm{ABC}$ be isosceles where $\mathrm{AB}=\mathrm{AC}$ and $\angle \mathrm{B}=\angle \mathrm{C}$

Given: Vertex angle $\mathrm{A}$ is twice the sum of the base angles $\mathrm{B}$ and $\mathrm{C}$. i.e., $\angle \mathrm{A}=2(\angle \mathrm{B}+\angle \mathrm{C})$

$\angle A=2(\angle B+\angle B)$

$\angle A=2(2 \angle B)$

$\angle A=4(\angle B)$

Now, We know that sum of angles in a triangle $=180^{\circ}$

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

$4 \angle B+\angle B+\angle B=180^{\circ}$

$6 \angle B=180^{\circ}$

$\angle B=30^{\circ}$

Since, $\angle B=\angle C$

$\angle \mathrm{B}=\angle \mathrm{C}=30^{\circ}$

And $\angle A=4 \angle B$

$\angle A=4 \times 30^{\circ}=120^{\circ}$

Therefore, angles of the given triangle are $30^{\circ}$ and $30^{\circ}$ and $120^{\circ}$.

Question 4: $\mathrm{PQR}$ is a triangle in which $\mathrm{PQ}=\mathrm{PR}$ and is any point on the side $\mathrm{PQ}$. Through $\mathrm{S}$, a line is drawn parallel to $Q R$ and intersecting $P R$ at $T$. Prove that $P S=P T$.

Solution: Given that $\mathrm{PQR}$ is a triangle such that $\mathrm{PQ}=\mathrm{PR}$ and $\mathrm{S}$ is any point on the side $\mathrm{PQ}$ and $\mathrm{ST}$ ॥ QR.

To prove: $\mathrm{PS}=\mathrm{PT}$

Since, $\mathrm{PQ}=\mathrm{PR}$, so $\triangle \mathrm{PQR}$ is an isosceles triangle.

$\angle P Q R=\angle P R Q$

Now, $\angle P S T=\angle P Q R$ and $\angle P T S=\angle P R Q$

[Corresponding angles as ST parallel to $\mathrm{QR}]$

Since, $\angle \mathrm{PQR}=\angle \mathrm{PRQ}$

$\angle \mathrm{PST}=\angle \mathrm{PTS}$

In $\triangle$ PST,

$\angle \mathrm{PST}=\angle \mathrm{PTS}$

$\triangle \mathrm{PST}$ is an isosceles triangle.

Therefore, $\mathrm{PS}=\mathrm{PT}$

Hence proved.

Exercise $10.4$

Question 1: In figure, It is given that $A B=C D$ and $A D=B C$. Prove that $\triangle A D C \cong \Delta C B A$.

Solution:

From figure, $A B=C D$ and $A D=B C$

To prove: $\triangle \mathrm{ADC} \cong \Delta \mathrm{CBA}$

Consider $\triangle \mathrm{ADC}$ and $\triangle \mathrm{CBA}$.

$\mathrm{AB}=\mathrm{CD}[\mathrm{Given}]$

$\mathrm{BC}=\mathrm{AD}[\mathrm{Given}]$

And $\mathrm{AC}=$ AC [Common side]

So, by SSS congruence criterion, we have

$\triangle \mathrm{ADC} \cong \Delta \mathrm{CBA}$

Hence proved.

Question 2: In a $\Delta \mathrm{PQR}$, if $\mathrm{PQ}=\mathrm{QR}$ and $\mathrm{L}, \mathrm{M}$ and $\mathrm{N}$ are the mid-points of the sides $\mathrm{PQ}, \mathrm{QR}$ and RP respectively. Prove that LN = MN.

Solution: Given: In $\triangle P Q R, P Q=Q R$ and $L, M$ and $N$ are the mid-points of the sides $P Q, Q R$ and $R P$ respectively

To prove: $L N=M N$

Join $\mathrm{L}$ and $\mathrm{M}, \mathrm{M}$ and $\mathrm{N}, \mathrm{N}$ and $\mathrm{L}$

We have $\mathrm{PL}=\mathrm{LQ}, \mathrm{QM}=\mathrm{MR}$ and $\mathrm{RN}=\mathrm{NP}$

[Since, $L, M$ and $N$ are mid-points of $P Q, Q R$ and $R P$ respectively] And also $\mathrm{PQ}=\mathrm{QR}$

$P L=L Q=Q M=M R=P N=L R \ldots \ldots$(i)

[ Using mid-point theorem]

$\mathrm{MN} \| \mathrm{PQ}$ and $\mathrm{MN}=\mathrm{PQ} / 2$

$\mathrm{MN}=\mathrm{PL}=\mathrm{LQ} \ldots \ldots$(ii)

Similarly, we have

$\mathrm{LN} \| \mathrm{QR}$ and $\mathrm{LN}=(1 / 2) \mathrm{QR}$

$\mathrm{LN}=\mathrm{QM}=\mathrm{MR} \ldots \ldots$(iii)

From equation (i), (ii) and (iii), we have

$\mathrm{PL}=\mathrm{LQ}=\mathrm{QM}=\mathrm{MR}=\mathrm{MN}=\mathrm{LN}$

This implies, $L N=M N$

Hence Proved.

Exercise $10.5$

Question 1: $A B C$ is a triangle and $D$ is the mid-point of $B C$. The perpendiculars from $D$ to $A B$ and $\mathrm{AC}$ are equal. Prove that the triangle is isosceles.

Solution: Given: $\mathrm{D}$ is the midpoint of $\mathrm{BC}$ and $\mathrm{PD}=\mathrm{DQ}$ in a triangle $\mathrm{ABC}$.

To prove: $A B C$ is isosceles triangle.

In $\triangle B D P$ and $\triangle C D Q$

$\mathrm{PD}=\mathrm{QD}$ (Given)

$B D=D C(D$ is mid-point $)$

$\angle B P D=\angle C Q D=90^{\circ}$

By RHS Criterion: $\triangle \mathrm{BDP} \cong \triangle \mathrm{CDQ}$

$\mathrm{BP}=\mathrm{CQ} \ldots$ (i) (By $\mathrm{CPCT})$

In $\triangle \mathrm{APD}$ and $\triangle \mathrm{AQD}$

$\mathrm{PD}=\mathrm{QD}$ (given)

$\mathrm{AD}=\mathrm{AD}(\mathrm{common})$

$\mathrm{APD}=\mathrm{AQD}=90^{\circ}$

By RHS Criterion: $\triangle \mathrm{APD} \cong \triangle \mathrm{AQD}$

So, $P A=Q A \ldots$ (ii) $($ By $C P C T)$

$\mathrm{BP}+\mathrm{PA}=\mathrm{CQ}+\mathrm{QA}$

$\mathrm{AB}=\mathrm{AC}$

Two sides of the triangle are equal, so $\mathrm{ABC}$ is an isosceles.

Question 2: $A B C$ is a triangle in which BE and CF are, respectively, the perpendiculars to the sides $A C$ and $A B$. If $B E=C F$, prove that $\Delta A B C$ is isosceles

Solution:

$\mathrm{ABC}$ is a triangle in which $\mathrm{BE}$ and $\mathrm{CF}$ are perpendicular to the sides $\mathrm{AC}$ and $\mathrm{AB}$ respectively s.t. BE $=\mathrm{CF}$.

To prove: $\triangle \mathrm{ABC}$ is isosceles

In $\triangle \mathrm{BCF}$ and $\triangle \mathrm{CBE}$,

$\angle B F C=C E B=90^{\circ}$ [Given]

$\mathrm{BC}=\mathrm{CB}$ [Common side]

And $\mathrm{CF}=\mathrm{BE}$ [Given]

By RHS congruence criterion: $\triangle \mathrm{BFC} \cong \Delta \mathrm{CEB}$

So, $\angle \mathrm{FBC}=\angle \mathrm{EBC}[\mathrm{By} \mathrm{CPCT}]$

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

$A C=A B$ [Opposite sides to equal angles are equal in a triangle]

Two sides of triangle $A B C$ are equal.

Therefore, $\triangle \mathrm{ABC}$ is isosceles. Hence Proved.

Question 3: If perpendiculars from any point within an angle on its arms are congruent. Prove that it lies on the bisector of that angle.

Solution:

Consider an angle $\mathrm{ABC}$ and $\mathrm{BP}$ be one of the arm within the angle.

Draw perpendiculars PN and PM on the arms BC and BA.

In $\triangle \mathrm{BPM}$ and $\triangle \mathrm{BPN}$,

$\angle \mathrm{BMP}=\angle \mathrm{BNP}=90^{\circ}$ [given]

$\mathrm{BP}=\mathrm{BP} \text { [Common side] }$

MP = NP [given]

By RHS congruence criterion: $\triangle \mathrm{BPM} \cong \triangle \mathrm{BPN}$

So, $\angle M B P=\angle N B P$ [ By CPCT]

BP is the angular bisector of $\angle A B C$.

Hence proved

Exercise $10.6$ Page No: $10.66$

Question 1: In $\triangle \mathrm{ABC}$, if $\angle \mathrm{A}=40^{\circ}$ and $\angle \mathrm{B}=60^{\circ} .$ Determine the longest and shortest sides of the triangle.

Solution: In $\triangle \mathrm{ABC}, \angle \mathrm{A}=40^{\circ}$ and $\angle \mathrm{B}=60^{\circ}$

We know, sum of angles in a triangle $=180^{\circ}$

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

$40^{\circ}+60^{\circ}+\angle C=180^{\circ}$

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

$\angle C=80^{\circ}$

Now, $40^{\circ}<60^{\circ}<80^{\circ}$

$=>A<\angle B<\angle C$

$=>\angle C$ is greater angle and $\angle A$ is smaller angle.

Now, $\angle A<\angle B<\angle C$

$\Rightarrow \angle A<\angle B<\angle C$

$=>\angle C$ is greater angle and $\angle A$ is smaller angle.

Now, $\angle A<\angle B<\angle C$

We know, side opposite to greater angle is larger and side opposite to smaller angle is smaller.

Therefore, AB is greater than AC, AC is greater than BC

$\mathrm{AB}$ is longest and $\mathrm{BC}$ is shortest side.

Question 2: In a $\triangle \mathrm{ABC}$, if $\angle \mathrm{B}=\angle \mathrm{C}=45^{\circ}$, which is the longest side?

Solution: $\ln \triangle \mathrm{ABC}, \angle \mathrm{B}=\angle \mathrm{C}=45^{\circ}$

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

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

$\angle A+45^{\circ}+45^{\circ}=180^{\circ}$

$\angle A=180^{\circ}-\left(45^{\circ}+45^{\circ}\right)$

$=180^{\circ}-90^{\circ}=90^{\circ}$

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

$\Rightarrow \angle B=\angle C<\angle A$

Therefore, $B C$ is the longest side.

Question 3: In $\triangle \mathrm{ABC}$, side $\mathrm{AB}$ is produced to $\mathrm{D}$ so that $\mathrm{BD}=\mathrm{BC}$. If $\angle \mathrm{B}=60^{\circ}$ and $\angle \mathrm{A}=70^{\circ}$.

Prove that: (i) $A D>C D$ (ii) $A D>A C$

Solution: $\ln \triangle \mathrm{ABC}$, side $\mathrm{AB}$ is produced to $\mathrm{D}$ so that $\mathrm{BD}=\mathrm{BC}$.

$\angle B=60^{\circ}$, and $\angle A=70^{\circ}$

To prove: (i) $A D>C D$ (ii) $A D>A C$

Construction: Join C and D

We know, sum of angles in a triangle $=180^{\circ}$

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

$70^{\circ}+60^{\circ}+\angle C=180^{\circ}$

$\angle C=180^{\circ}-\left(130^{\circ}\right)=50^{\circ}$

$\angle C=50^{\circ}$

$\angle A C B=50^{\circ} \ldots \ldots($ i $)$

And also in $\triangle B D C$

$\angle \mathrm{DBC}=180^{\circ}-\angle \mathrm{ABC}=180-60^{\circ}=120^{\circ}$

[\angleDBA is a straight line]

and $B D=B C$ [given]

$\angle B C D=\angle B D C$ [Angles opposite to equal sides are equal]

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

$\angle \mathrm{DBC}+\angle \mathrm{BCD}+\angle \mathrm{BDC}=180^{\circ}$

$120^{\circ}+\angle B C D+\angle B C D=180^{\circ}$

$120^{\circ}+2 \angle \mathrm{BCD}=180^{\circ}$

$2 \angle \mathrm{BCD}=180^{\circ}-120^{\circ}=60^{\circ}$

$\angle B C D=30^{\circ}$

$\angle B C D=\angle B D C=30^{\circ} \ldots$(ii)

Now, consider $\triangle$ ADC.

$\angle \mathrm{DAC}=70^{\circ}$ [given]

$\angle A D C=30^{\circ}$ [From (ii)]

$\angle A C D=\angle A C B+\angle B C D=50^{\circ}+30^{\circ}$

$=80^{\circ}$ [From (i) and (ii)]

Now, $\angle A D C<\angle D A C<\angle A C D$

$A C [Side opposite to greater angle is longer and smaller angle is smaller]$A D>C D$and$A D>A C$Hence proved. Question 4: Is it possible to draw a triangle with sides of length$2 \mathrm{~cm}, 3 \mathrm{~cm}$and$7 \mathrm{~cm}$? Solution: Lengths of sides are$2 \mathrm{~cm}, 3 \mathrm{~cm}$and$7 \mathrm{~cm}$. A triangle can be drawn only when the sum of any two sides is greater than the third side. So, let's check the rule.$2+3 \not>7$or$2+3<72+7>3$and$3+7>2$Here$2+3 \not>7$So, the triangle does not exit. Exercise VSAQs Question 1 : In two congruent triangles$\mathrm{ABC}$and$\mathrm{DEF}$, if$\mathrm{AB}=\mathrm{DE}$and$\mathrm{BC}=\mathrm{EF}$. Name the pairs of equal angles. Solution: In two congruent triangles$A B C$and DEF, if$A B=D E$and$B C=E F$, then$\angle A=\angle D, \angle B=\angle E$and$\angle C=\angle F$Question 2: In two triangles$A B C$and DEF, it is given that$\angle A=\angle D, \angle B=\angle E$and$\angle C=\angle F$. Are the two triangles necessarily congruent? Solution: No. Reason: Two triangles are not necessarily congruent, because we know only angle-angle-angle (AAA) criterion. This criterion can produce similar but not congruent triangles. Question 3: If$A B C$and DEF are two triangles such that$A C=2.5 \mathrm{~cm}, B C=5 \mathrm{~cm}, C=75^{\circ}$, DE =$2.5 \mathrm{~cm}, \mathrm{DF}=5 \mathrm{~cm}$and$\mathrm{D}=75^{\circ}$. Are two triangles congruent? Solution: Yes. Reason: Given triangles are congruent as$\mathrm{AC}=\mathrm{DE}=2.5 \mathrm{~cm}, \mathrm{BC}=\mathrm{DF}=5 \mathrm{~cm}$and$\angle D=\angle C=75^{\circ}$. By SAS theorem triangle$\mathrm{ABC}$is congruent to triangle EDF. Question 4: In two triangles$A B C$and$A D C$, if$A B=A D$and$B C=C D$. Are they congruent? Solution: Yes. Reason: Given triangles are congruent as$\mathrm{AB}=\mathrm{AD}\mathrm{BC}=\mathrm{CD}$and AC [ common side] By SSS theorem triangle$A B C$is congruent to triangle$A D C$. Question 5: In triangles$A B C$and$C D E$, if$A C=C E, B C=C D, \angle A=60^{\circ}, \angle C=30^{\circ}$and$\angle D=90^{\circ}$Are two triangles congruent? Solution: Yes. Reason: Given triangles are congruent Here$A C=C EB C=C D\angle B=\angle D=90^{\circ}$By SSA criteria triangle$A B C$is congruent to triangle$C D E .$Question 6:$\mathrm{ABC}$is an isosceles triangle in which$\mathrm{AB}=$AC. BE and CF are its two medians. Show that$B E=C F$Solution:$A B C$is an isosceles triangle (given)$\mathrm{AB}=\mathrm{AC}$(given) BE and CF are two medians (given) To prove:$\mathrm{BE}=\mathrm{CF}$In$\triangle \mathrm{CFB}$and$\triangle \mathrm{BEC}\mathrm{CE}=\mathrm{BF}($Since,$\mathrm{AC}=\mathrm{AB}=A C / 2=A B / 2=C E=B F)\mathrm{BC}=\mathrm{BC}$(Common)$\angle \mathrm{ECB}=\angle \mathrm{FBC}$(Angle opposite to equal sides are equal) By SAS theorem:$\triangle \mathrm{CFB} \cong \triangle \mathrm{BEC}$So,$B E=C F\$ (By c.p.c.t)