Exercise 7.1 (Revised) - Chapter 7 - Triangles - Ncert Solutions class 9 - Maths

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Chapter 7 - Triangles | NCERT Solutions for Class 9 Maths

Ex 7.1 Question 1.

In quadrilateral $A B C D$ (See figure). $A C=A D$ and $A B$ bisects $\angle A$. Show that $\triangle A B C \cong \triangle A B D$. What can you say about BC and BD?

Answer.

Given: In quadrilateral $\mathrm{ABCD}, \mathrm{AC}=\mathrm{AD}$ and $\mathrm{AB}$ bisects $\angle \mathrm{A}$.

To prove: $\triangle \mathrm{ABC} \cong \triangle \mathrm{ABD}$

Proof: In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{ABD}$,
$\mathrm{AC}=\mathrm{AD}[$ Given $]$
$\angle \mathrm{BAC}=\angle \mathrm{BAD}[\because$ AB bisects $\angle \mathrm{A}]$
$\mathrm{AB}=\mathrm{AB}[$ Common $]$
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{ABD}$ [By SAS congruency]

Thus $\mathrm{BC}=\mathrm{BD}$ [By C.P.C.T.]
Ex 7.1 Question 2.

$\mathrm{ABCD}$ is a quadrilateral in which $\mathrm{AD}=\mathrm{BC}$ and $\angle \mathrm{DAB}=\angle \mathrm{CBA}$. (See figure). Prove that:

(i) $\triangle$ ABD $\cong \triangle B A C$
(ii) $B D=A C$
(iii) $\angle \mathrm{ABD}=\angle \mathrm{BAC}$

Answer.

(i) In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{ABD}$,
$\mathrm{BC}=\mathrm{AD}[$ Given $]$
$\angle \mathrm{DAB}=\angle \mathrm{CBA}[$ Given]
$\mathrm{AB}=\mathrm{AB}[$ Common]
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{ABD}$ [By SAS congruency]

Thus AC $=$ BD [By C.P.C.T.]
(ii) Since $\triangle A B C \cong \triangle A B D$
$\therefore A C=B D[B y$ C.P.C.T.]
(iii) Since $\triangle \mathrm{ABC} \cong \triangle \mathrm{ABD}$
$\therefore \angle \mathrm{ABD}=\angle \mathrm{BAC}$ [By C.P.C.T.]

Ex 7.1 Question 3.

AD and BC are equal perpendiculars to a line segment $A B$. Show that CD bisects $A B$ (See figure)

Answer.

In $\triangle B O C$ and $\triangle A O D$,
$\angle \mathrm{OBC}=\angle \mathrm{OAD}=90^{\circ}$ [Given]
$\angle \mathrm{BOC}=\angle \mathrm{AOD}$ [Vertically Opposite angles]
$\mathrm{BC}=\mathrm{AD}[$ Given $]$
$\therefore \triangle \mathrm{BOC} \cong \triangle \mathrm{AOD}$ [By ASA congruency]
$\Rightarrow \mathrm{OB}=\mathrm{OA}$ and $\mathrm{OC}=\mathrm{OD}[$ By C.P.C.T. $]$

Ex 7.1 Question 4.

$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (See figure). Show that $\triangle A B C \cong \triangle C D A$.

Answer.

AC being a transversal. [Given]

Therefore $\angle \mathrm{DAC}=\angle \mathrm{ACB}$ [Alternate angles]

Now $p \| q$ [Given]

And $\mathrm{AC}$ being a transversal. [Given]

Therefore $\angle \mathrm{BAC}=\angle \mathrm{ACD}$ [Alternate angles]

Now in $\triangle \mathrm{ABC}$ and $\triangle \mathrm{ADC}$,
$\angle \mathrm{ACB}=\angle \mathrm{DAC}$ [Proved above]
$\angle \mathrm{BAC}=\angle \mathrm{ACD}$ [Proved above]
$\mathrm{AC}=\mathrm{AC}[$ Common $]$
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{CDA}$ [By ASA congruency]

Ex 7.1 Question 5.

Line ${ }^l$ is the bisector of the angle $\mathrm{A}$ and $\mathrm{B}$ is any point on ${ }^l$. BP and BQ are perpendiculars from $B$ to the arms of $\angle \mathbf{A}$. Show that:

(i) $\triangle \mathrm{APB} \cong \triangle A Q B$
(ii) $\mathrm{BP}=\mathrm{BQ}$ or $\mathrm{P}$ is equidistant from the arms of $\angle \mathrm{A}$ (See figure).

Answer.

Given: Line ${ }^l$ bisects $\angle \mathrm{A}$.
$
\therefore \angle \mathrm{BAP}=\angle \mathrm{BAQ}
$
(i) In $\triangle \mathrm{ABP}$ and $\triangle \mathrm{ABQ}$,
$
\begin{aligned}
& \angle \mathrm{BAP}=\angle \mathrm{BAQ} \text { [Given] } \\
& \angle \mathrm{BPA}=\angle \mathrm{BQA}=90^{\circ} \text { [Given] } \\
& \mathrm{AB}=\mathrm{AB} \text { [Common] } \\
& \therefore \triangle \mathrm{APB} \cong \triangle \mathrm{AQB} \text { [By ASA congruency] }
\end{aligned}
$
(ii) Since $\triangle \mathrm{APB} \cong \triangle \mathrm{AQB}$
$
\therefore B P=B Q[B y \text { C.P.C.T. }]
$
$\Rightarrow \mathrm{B}$ is equidistant from the arms of $\angle \mathrm{A}$.

Ex 7.1 Question 6.

$\text {In figure, } A C=A B, A B=A D \text { and } \angle B A D=\angle E A C \text {. Show that } B C=D E . \wedge$

Answer.

Given that $\angle \mathrm{BAD}=\angle \mathrm{EAC}$

Adding $\angle$ DAC on both sides, we get
$
\begin{aligned}
& \angle \mathrm{BAD}+\angle \mathrm{DAC}=\angle \mathrm{EAC}+\angle \mathrm{DAC} \\
& \Rightarrow \angle \mathrm{BAC}=\angle \mathrm{EAD} \ldots \ldots \ldots . \text { (i) }
\end{aligned}
$

Now in $\triangle \mathrm{ABC}$ and $\triangle \mathrm{AED}$,
$\mathrm{AB}=\mathrm{AD}[$ Given $]$
$\mathrm{AC}=\mathrm{AE}[$ Given $]$
$\angle \mathrm{BAC}=\angle \mathrm{DAE}$ [From eq. (i)]
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{ADE}$ [By SAS congruency]
$\Rightarrow \mathrm{BC}=\mathrm{DE}[\mathrm{By}$ C.P.C.T.]

Ex 7.1 Question 7.

$A B$ is a line segment and $P$ is the mid-point. $D$ and $E$ are points on the same side of $A B$ such that $\angle B A D=\angle A B E$ and $\angle E P A=\angle D P B$. Show that:
(i) $\triangle$ DAF $\cong \triangle F B P$
(ii) $A D=B E$ (See figure)

Answer.

Given that $\angle \mathrm{EPA}=\angle \mathrm{DPB}$

Adding $\angle$ EPD on both sides, we get
$
\begin{aligned}
& \angle \mathrm{EPA}+\angle \mathrm{EPD}=\angle \mathrm{DPB}+\angle \mathrm{EPD} \\
& \Rightarrow \angle \mathrm{APD}=\angle \mathrm{BPE} \ldots \ldots \ldots \text { (i) }
\end{aligned}
$

Now in $\triangle \mathrm{APD}$ and $\triangle \mathrm{BPE}$,
$
\begin{aligned}
& \angle \mathrm{PAD}=\angle \mathrm{PBE}[\because \angle \mathrm{BAD}=\angle \mathrm{ABE} \text { (given), } \\
& \therefore \angle \mathrm{PAD}=\angle \mathrm{PBE}]
\end{aligned}
$
$A P=P B[P$ is the mid-point of $A B]$
$
\begin{aligned}
& \angle \mathrm{APD}=\angle \mathrm{BPE} \text { [From eq. (i)] } \\
& \therefore \triangle \mathrm{DPA} \cong \triangle \mathrm{EBP}[\text { By ASA congruency] } \\
& \Rightarrow \mathrm{AD}=\mathrm{BE} \text { [ By C.P.C.T.] }
\end{aligned}
$

Ex 7.1 Question 8.

In right triangle $A B C$, right angled at $C, M$ is the mid-point of hypotenuse $A B$. $C$ is joined to $M$ and produced to a point $D$ such that $D M=C M$. Point $D$ is joined to point $B$. (See figure)

Show that:
(i) $\triangle$ AMC $\cong \triangle$ BMD
(ii) $\angle \mathrm{DBC}$ is a right angle.
(iii) $\triangle \mathrm{DBC} \cong \triangle \mathrm{ACB}$
(iv) $C M=\frac{1}{2} \mathrm{AB}$

Answer.

(i) In $\triangle \mathrm{AMC}$ and $\triangle \mathrm{BMD}$,
$A M=B M[A B$ is the mid-point of $A B]$
$\angle \mathrm{AMC}=\angle \mathrm{BMD}$ [Vertically opposite angles]
$\mathrm{CM}=\mathrm{DM}[$ Given]
$\therefore \triangle \mathrm{AMC} \cong \triangle \mathrm{BMD}$ [By SAS congruency]
$\therefore \angle \mathrm{ACM}=\angle \mathrm{BDM}$
$\angle \mathrm{CAM}=\angle \mathrm{DBM}$ and $\mathrm{AC}=\mathrm{BD}$ [By C.P.C.T.]
(ii) For two lines $\mathrm{AC}$ and $\mathrm{DB}$ and transversal $\mathrm{DC}$, we have,
$\angle \mathrm{ACD}=\angle \mathrm{BDC}$ [Alternate angles]

$\therefore A C \| D B$

Now for parallel lines $\mathrm{AC}$ and $\mathrm{DB}$ and for transversal BC.
$\angle \mathrm{DBC}=\angle \mathrm{ACB}$ [Alternate angles]

But $\triangle \mathrm{ABC}$ is a right angled triangle, right angled at $\mathrm{C}$.
$
\therefore \angle \mathrm{ACB}=90^{\circ}
$

Therefore $\angle \mathrm{DBC}=90^{\circ}$ [Using eq. (ii) and (iii)]
$\Rightarrow \angle \mathrm{DBC}$ is a right angle.
(iii) Now in $\triangle \mathrm{DBC}$ and $\triangle \mathrm{ABC}$,
$D B=A C[$ Proved in part (i)]
$\angle \mathrm{DBC}=\angle \mathrm{ACB}=90^{\circ}$ [Proved in part (ii)]
$\mathrm{BC}=\mathrm{BC}[$ Common $]$
$\therefore \triangle \mathrm{DBC} \cong \triangle \mathrm{ACB}$ [By SAS congruency]
(iv) Since $\triangle \mathrm{DBC} \cong \triangle \mathrm{ACB}$ [Proved above]
$\therefore \mathrm{DC}=\mathrm{AB}$

$\begin{aligned}
& \Rightarrow A M+C M=A B \\
& \Rightarrow C M+C M=A B[\because D M=C M] \\
& \Rightarrow 2 C M=A B \\
& \Rightarrow C M=\frac{1}{2} A B
\end{aligned}$