Exercise 7.3 (Revised) - Chapter 7 - Triangles - Ncert Solutions class 9 - Maths

Share this to Friend on WhatsApp

Chapter 7 - Triangles | NCERT Solutions for Class 9 Maths

Ex 7.3 Question 1.

$\triangle A B C$ and $\triangle D B C$ are two isosceles triangles on the same base $B C$ and vertices $A$ and $D$ are on the same side of BC (See figure). If AD is extended to intersect BC at $P$, show that:

(i) $\triangle \mathrm{ABD} \cong \triangle \mathrm{ACD}$
(ii) $\triangle \mathbf{A B P} \cong \triangle \mathrm{ACP}$
(iii) AP bisects $\angle \mathbf{A}$ as well as $\angle \mathbf{D}$.
(iv) AP is the perpendicular bisector of BC.

Answer.

(i) $\triangle \mathrm{ABC}$ is an isosceles triangle.
$
\therefore A B=A C
$
$\triangle \mathrm{DBC}$ is an isosceles triangle.
$
\therefore B D=C D
$

Now in $\triangle \mathrm{ABD}$ and $\triangle \mathrm{ACD}$,
$A B=A C[$ Given $]$
$B D=C D[$ Given $]$
$\mathrm{AD}=\mathrm{AD}[$ Common $]$
$\therefore \Delta \mathrm{ABD} \cong \triangle \mathrm{ACD}$ [By SSS congruency]
$\Rightarrow \angle \mathrm{BAD}=\angle \mathrm{CAD}$ [By C.P.C.T.]

(ii) Now in $\triangle \mathrm{ABP}$ and $\triangle \mathrm{ACP}$,
$A B=A C[$ Given $]$
$\angle \mathrm{BAD}=\angle \mathrm{CAD}[$ From eq. (i)]
$\mathrm{AP}=\mathrm{AP}$
$\therefore \Delta \mathrm{ABP} \cong \triangle \mathrm{ACP}$ [By SAS congruency]
(iii) Since $\triangle \mathrm{ABP} \cong \triangle \mathrm{ACP}$ [From part (ii)]
$\Rightarrow \angle \mathrm{BAP}=\angle \mathrm{CAP}[\mathrm{By}$ C.P.C.T.]
$\Rightarrow$ AP bisects $\angle \mathrm{A}$.

Since $\triangle \mathrm{ABD} \cong \triangle \mathrm{ACD}[$ From part (i) $]$
$\Rightarrow \angle \mathrm{ADB}=\angle \mathrm{ADC}$ [By C.P.C.T.]

Now $\angle \mathrm{ADB}+\angle \mathrm{BDP}=180^{\circ}$ [Linear pair]

And $\angle \mathrm{ADC}+\angle \mathrm{CDP}=180^{\circ}$ [Linear pair]

From eq. (iii) and (iv),
$
\angle \mathrm{ADB}+\angle \mathrm{BDP}=\angle \mathrm{ADC}+\angle \mathrm{CDP}
$

$
\begin{aligned}
& \Rightarrow \angle \mathrm{ADB}+\angle \mathrm{BDP}=\angle \mathrm{ADB}+\angle \mathrm{CDP}[\text { Using (ii)] } \\
& \Rightarrow \angle \mathrm{BDP}=\angle \mathrm{CDP}
\end{aligned}
$
$\Rightarrow$ DP bisects $\angle \mathrm{D}$ or AP bisects $\angle \mathrm{D}$.
(iv) Since $\triangle A B P \cong \triangle A C P[$ From part (ii)]
$
\therefore B P=P C[B y \text { C.P.C.T. }]
$

And $\angle \mathrm{APB}=\angle \mathrm{APC}[\mathrm{By}$ C.P.C.T. $]$ $\qquad$ (vi)

Now $\angle \mathrm{APB}+\angle \mathrm{APC}=180^{\circ}$ [Linear pair $]$
$
\begin{aligned}
& \Rightarrow \angle \mathrm{APB}+\angle \mathrm{APC}=180^{\circ} \text { [Using eq. (vi)] } \\
& \Rightarrow 2 \angle \mathrm{APB}=180^{\circ} \\
& \Rightarrow \angle \mathrm{APB}=90^{\circ}
\end{aligned}
$
$
\Rightarrow \mathrm{AP} \perp \mathrm{BC}
$

From eq. (v), we have BP PC and from (vii), we have proved AP $\perp$. So, collectively AP is perpendicular bisector of $B C$.

Ex 7.3 Question 2.

$A D$ is an altitude of an isosceles triangle $A B C$ in which $A B=A C$. Show that:
(i) AD bisects BC.
(ii) AD bisects $\angle \mathrm{A}$.

Answer.

In $\triangle A B D$ and $\triangle A C D$,
$A B=A C[$ Given]
$
\angle \mathrm{ADB}=\angle \mathrm{ADC}=90^{\circ}[\mathrm{AD} \perp \mathrm{BC}]
$

$\mathrm{AD}=\mathrm{AD}$ [Common]
$\therefore \triangle \mathrm{ABD} \cong \triangle \mathrm{ACD}[\mathrm{RHS}$ rule of congruency]
$\Rightarrow B D=D C$ [By C.P.C.T.]
$\Rightarrow A D$ bisects $B C$

Also $\angle \mathrm{BAD}=\angle \mathrm{CAD}$ [By C.P.C.T.]
$\Rightarrow$ AD bisects $\angle \mathrm{A}$.

Ex 7.3 Question 3.

Two sides $A B$ and $B C$ and median $A M$ of the triangle $A B C$ are respectively equal to side $P Q$ and QR and median PN of $\triangle$ PQR (See figure). Show that:

(i) $\triangle$ ABM $\cong \triangle$ PQN
(ii) $\triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}$

Answer.

AM is the median of $\triangle \mathrm{ABC}$.
$
\therefore B M=M C=\frac{1}{2} B C
$
$\mathrm{PN}$ is the median of $\triangle \mathrm{PQR}$.

Now $\mathrm{BC}=\mathrm{QR}\left[\right.$ Given] $\Rightarrow \frac{1}{2} \mathrm{BC}=\frac{1}{2} \mathrm{QR}$
$
\therefore B M=Q N
$
(i) Now in $\triangle \mathrm{ABM}$ and $\triangle \mathrm{PQN}$,
$A B=P Q[$ Given $]$
$\mathrm{AM}=\mathrm{PN}[$ Given $]$
$\mathrm{BM}=\mathrm{QN}[$ From eq. (iii)]
$\therefore \triangle \mathrm{ABM} \cong \triangle \mathrm{PQN}$ [By SSS congruency]

$
\Rightarrow \angle \mathrm{B}=\angle \mathrm{Q} \text { [By C.P.C.T.] }
$
(ii) In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$,
$\mathrm{AB}=\mathrm{PQ}$ [Given]
$\angle \mathrm{B}=\angle \mathrm{Q}$ [Prove above]
$\mathrm{BC}=\mathrm{QR}[$ Given]
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}$ [By SAS congruency]

Ex 7.3 Question 4.

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Answer.

In $\triangle B E C$ and $\triangle C F B$,

$
\begin{aligned}
& \angle \mathrm{BEC}=\angle \mathrm{CFB}\left[\text { Each } 90^{\circ}\right] \\
& \mathrm{BC}=\mathrm{BC} \text { [Common] } \\
& \mathrm{BE}=\mathrm{CF} \text { [Given] } \\
& \therefore \Delta \mathrm{BEC} \cong \Delta \mathrm{CFB} \text { [RHS congruency] } \\
& \Rightarrow \mathrm{EC}=\mathrm{FB} \text { [By C.P.C.T.] ....(i) }
\end{aligned}
$

Now In $\triangle \mathrm{AEB}$ and $\triangle \mathrm{AFC}$
$
\begin{aligned}
& \angle \mathrm{AEB}=\angle \mathrm{AFC}\left[\text { Each } 90^{\circ}\right. \text { ] } \\
& \angle \mathrm{A}=\angle \mathrm{A}[\text { Common] } \\
& \mathrm{BE}=\mathrm{CF} \text { [Given] }
\end{aligned}
$

$\therefore \Delta \mathrm{AEB} \cong \Delta \mathrm{AFC}$ [ASA congruency]
$\Rightarrow \mathrm{AE}=\mathrm{AF}[\mathrm{By}$ C.P.C.T.] $\qquad$
Adding eq. (i) and (ii), we get,
$E C+A E=F B+A F$
$\Rightarrow \mathrm{AB}=\mathrm{AC}$
$\Rightarrow A B C$ is an isosceles triangle.

Ex 7.3 Question 5.

$\mathrm{ABC}$ is an isosceles triangles with $\mathrm{AB}=\mathrm{AC}$. Draw $\mathrm{AP} \perp \mathrm{BC}$ and show that $\angle \mathrm{B}=\angle \mathrm{C}$.

Answer.

Given: $\mathrm{ABC}$ is an isosceles triangle in which $\mathrm{AB}=\mathrm{AC}$

To prove: $\angle \mathrm{B}=\angle \mathrm{C}$

Construction: Draw AP $\perp \mathrm{BC}$

Proof: In $\triangle \mathrm{ABP}$ and $\triangle \mathrm{ACP}$
$
\begin{aligned}
& \angle \mathrm{APB}=\angle \mathrm{APC}=90^{\circ} \text { [By construction] } \\
& \mathrm{AB}=\mathrm{AC} \text { [Given] } \\
& \mathrm{AP}=\mathrm{AP} \text { [Common] } \\
& \therefore \triangle \mathrm{ABP} \cong \triangle \mathrm{ACP} \text { [RHS congruency] } \\
& \Rightarrow \angle \mathrm{B}=\angle \mathrm{C} \text { [By C.P.C.T. }]
\end{aligned}
$