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

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Chapter 7 - Triangles | NCERT Solutions for Class 9 Maths

Example 1:

In Fig. 7.8, $\mathrm{OA}=\mathrm{OB}$ and $\mathrm{OD}=\mathrm{OC}$. Show that
(i) $\triangle \mathrm{AOD} \cong \triangle \mathrm{BOC}$ and
(ii) $\mathrm{AD} \| \mathrm{BC}$.

Solution :

(i) You may observe that in $\triangle \mathrm{AOD}$ and $\triangle \mathrm{BOC}$,
$
\left.\begin{array}{l}
\mathrm{OA}=\mathrm{OB} \\
\mathrm{OD}=\mathrm{OC}
\end{array}\right\}
$

Also, since $\angle \mathrm{AOD}$ and $\angle \mathrm{BOC}$ form a pair of vertically opposite angles, we have
$
\begin{aligned}
& \angle \mathrm{AOD}=\angle \mathrm{BOC} . \\
& \triangle \mathrm{AOD} \cong \triangle \mathrm{BOC}
\end{aligned}
$

So,
(by the SAS congruence rule)
(ii) In congruent triangles $\mathrm{AOD}$ and $\mathrm{BOC}$, the other corresponding parts are also equal.

So, $\angle \mathrm{OAD}=\angle \mathrm{OBC}$ and these form a pair of alternate angles for line segments $\mathrm{AD}$ and $\mathrm{BC}$.
Therefore,
$\mathrm{AD} \| \mathrm{BC}$.

Example 2:

AB is a line segment and line $l$ is its perpendicular bisector. If a point $\mathrm{P}$ lies on $l$, show that $\mathrm{P}$ is equidistant from $\mathrm{A}$ and $\mathrm{B}$.
Solution :

Line $l \perp \mathrm{AB}$ and passes through $\mathrm{C}$ which is the mid-point of $\mathrm{AB}$ (see Fig. 7.9). You have to show that $\mathrm{PA}=\mathrm{PB}$. Consider $\triangle \mathrm{PCA}$ and $\triangle \mathrm{PCB}$.
We have $\mathrm{AC}=\mathrm{BC} \quad(\mathrm{C}$ is the mid-point of $\mathrm{AB})$
$
\begin{aligned}
\angle \mathrm{PCA} & =\angle \mathrm{PCB}=90^{\circ} \\
\mathrm{PC} & =\mathrm{PC}
\end{aligned}
$

So, $\triangle \mathrm{PCA} \cong \triangle \mathrm{PCB}$ (SAS rule)
and so, $\mathrm{PA}=\mathrm{PB}$, as they are corresponding sides of congruent triangles.
(Common)
Now, let us construct two triangles, whose sides are $4 \mathrm{~cm}$ and $5 \mathrm{~cm}$ and one of the angles is $50^{\circ}$ and this angle is not included in between the equal sides (see Fig. 7.10). Are the two triangles congruent?

Notice that the two triangles are not congruent.
Repeat this activity with more pairs of triangles. You will observe that for triangles to be congruent, it is very important that the equal angles are included between the pairs of equal sides.
So, SAS congruence rule holds but not ASS or SSA rule.
Next, try to construct the two triangles in which two angles are $60^{\circ}$ and $45^{\circ}$ and the side included between these angles is $4 \mathrm{~cm}$ (see Fig. 7.11).

Cut out these triangles and place one triangle on the other. What do you observe? See that one triangle covers the other completely; that is, the two triangles are congruent. Repeat this activity with more pairs of triangles. You will observe that equality of two angles and the included side is sufficient for congruence of triangles.

This result is the Angle-Side-Angle criterion for congruence and is written as ASA criterion. You have verified this criterion in earlier classes, but let us state and prove this result.

Since this result can be proved, it is called a theorem and to prove it, we use the SAS axiom for congruence.

Example 3 :

Line-segment $\mathrm{AB}$ is parallel to another line-segment $\mathrm{CD}$. $\mathrm{O}$ is the mid-point of $A D$ (see Fig. 7.15). Show that (i) $\triangle A O B \cong \triangle D O C$ (ii) $O$ is also the mid-point of BC.

Solution :

(i) Consider $\triangle \mathrm{AOB}$ and $\triangle \mathrm{DOC}$.
$
\angle \mathrm{ABO}=\angle \mathrm{DCO}
$
(Alternate angles as $\mathrm{AB} \| \mathrm{CD}$ and $\mathrm{BC}$ is the transversal)
$
\angle \mathrm{AOB}=\angle \mathrm{DOC}
$
(Vertically opposite angles)
$
\mathrm{OA}=\mathrm{OD}
$
(Given)
Therefore, $\quad \triangle \mathrm{AOB} \cong \triangle \mathrm{DOC}$
(AAS rule)
(ii)
$
\mathrm{OB}=\mathrm{OC}
$
(CPCT)
So, $\mathrm{O}$ is the mid-point of $\mathrm{BC}$.

Example 4:

 In $\triangle \mathrm{ABC}$, the bisector $\mathrm{AD}$ of $\angle \mathrm{A}$ is perpendicular to side $\mathrm{BC}$ (see Fig. 7.27). Show that $A B=A C$ and $\triangle A B C$ is isosceles.
Solution :

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

Example 5 :

$\mathrm{E}$ and $\mathrm{F}$ are respectively the mid-points of equal sides $\mathrm{AB}$ and $\mathrm{AC}$ of $\triangle \mathrm{ABC}$ (see Fig. 7.28). Show that $\mathrm{BF}=\mathrm{CE}$.

Solution :

In $\triangle \mathrm{ABF}$ and $\triangle \mathrm{ACE}$,

Example 6 :

In an isosceles triangle $\mathrm{ABC}$ with $\mathrm{AB}=\mathrm{AC}, \mathrm{D}$ and $\mathrm{E}$ are points on $\mathrm{BC}$ such that $B E=C D$ (see Fig. 7.29). Show that $A D=A E$.
Solution :

In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{ACE}$,

Example 7:

AB is a line-segment. $\mathrm{P}$ and $\mathrm{Q}$ are points on opposite sides of $A B$ such that each of them is equidistant from the points $\mathrm{A}$ and $\mathrm{B}$ (see Fig. 7.37). Show that the line $\mathrm{PQ}$ is the perpendicular bisector of $\mathrm{AB}$.

Solution :

You are given that $\mathrm{PA}=\mathrm{PB}$ and $\mathrm{QA}=\mathrm{QB}$ and you are to show that $\mathrm{PQ} \perp \mathrm{AB}$ and $P Q$ bisects $A B$. Let $P Q$ intersect $A B$ at $C$.
Can you think of two congruent triangles in this figure?
Let us take $\Delta \mathrm{PAQ}$ and $\Delta \mathrm{PBQ}$.
In these triangles,

From (1) and (2), you can easily conclude that $\mathrm{PQ}$ is the perpendicular bisector of $\mathrm{AB}$.
[Note that, without showing the congruence of $\triangle \mathrm{PAQ}$ and $\triangle \mathrm{PBQ}$, you cannot show that $\triangle \mathrm{PAC} \cong \triangle \mathrm{PBC}$ even though $\mathrm{AP}=\mathrm{BP}$
(Given)
$
\mathrm{PC}=\mathrm{PC}
$
(Common)
and $\angle \mathrm{PAC}=\angle \mathrm{PBC}$ (Angles opposite to equal sides in $\triangle \mathrm{APB}$ )

It is because these results give us SSA rule which is not always valid or true for congruence of triangles. Also the angle is not included between the equal pairs of sides.]
Let us take some more examples.
Example 8 :

$\mathrm{P}$ is a point equidistant from two lines $l$ and $m$ intersecting at point $\mathrm{A}$ (see Fig. 7.38). Show that the line AP bisects the angle between them.
Solution :

You are given that lines $l$ and $m$ intersect each other at A. Let $\mathrm{PB} \perp l$, $\mathrm{PC} \perp m$. It is given that $\mathrm{PB}=\mathrm{PC}$.
You are to show that $\angle \mathrm{PAB}=\angle \mathrm{PAC}$.

Let us consider $\triangle \mathrm{PAB}$ and $\Delta \mathrm{PAC}$. In these two triangles,