NCERT Exemplar Class 9 Maths Solutions Chapter 6 Lines and Angles

Question 1: Write the correct answer in each of the following:
$\text{In Fig. 6.1, if }AB\parallel CD\parallel EF,PQ\parallel RS,$ $\mathrm{\angle }RQD={25}^{\circ }$ $\text{ and }\mathrm{\angle }CQP={60}^{\circ }$$\text{ and }\mathrm{\angle }CQP={60}^{\circ }\text{, then}$ $\mathrm{\angle }QRS\text{ is equal to }$

$(A){85}^{\circ }$

$(B){135}^{\circ }$

$(C){145}^{\circ }$
$(D){110}^{\circ }$

Answer:

Here in this question, it is given,
$PQ\parallel RS$
$\mathrm{\angle }PQC=\mathrm{\angle }BRQ={60}^{\circ }$ $\text{(Alternate Exterior angles)}$
$\mathrm{\angle }PQC={60}^{\circ }$
$\mathrm{\angle }DQR=\mathrm{\angle }QRA={25}^{\circ }$ $\text{(Alternate interior angles)}$
$\mathrm{\angle }QRS=\mathrm{\angle }QRA+\mathrm{\angle }ARS$
$=\mathrm{\angle }QRA+({180}^{\circ }-\mathrm{\angle }BRS)$ $\text{(Linear pair angle)}$
$={25}^{\circ }+{180}^{\circ }-{60}^{\circ }={205}^{\circ }-{60}^{\circ }={145}^{\circ }$

Hence, option C is the right answer.

Question:2 Write the correct answer in each of the following:
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(A) an isosceles triangle
(B) An obtuse triangle
(C) An equilateral triangle
(D) a right triangle

In a $\mathrm{△}ABC$

$\mathrm{\angle }A=\mathrm{\angle }B+\mathrm{\angle }C.....(i,)$

$\text{Now sum of all angles =}{180}^{\circ }$

$,\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$
$\mathrm{\angle }A+\mathrm{\angle }A={180}^{\circ }$ $\text{(from i)}$
$\mathrm{\angle }A={90}^{\circ }$

Thus triangle ABC is a right-angled triangle.

Hence Option (D) is the correct answer.

Question:3 Write the correct answer in each of the following:
An exterior angle of a triangle is 105°, and its two interior opposite angles are equal. Each of these equal angles is
$(A)$ $34{\frac{1}{2}}^{\circ }$
$(B)$ $52\frac{1}{t}he{2}^{\circ }$
$(C)$ $72{\frac{1}{2}}^{\circ }$
$(D)$ ${75}^{\circ }$

Answer:

Let the two equal interior angles be $x^{\circ }$.

The sum of opposite interior angles of a triangle = Exterior angle of the triangle.

$x^{\circ }+x^{\circ }={105}^{\circ }$

$2x^{\circ }={105}^{\circ }$

$x^{\circ }=52{\frac{1}{2}}^{\circ }$

SoAnoption (B) is the correct answer.

Question:4 Write the correct answer in each of the following: The angles of a triangle are in the ratio 5 : 3: 7. The triangle is
(A) an acute angled triangle
(B) an obtuse-angled triangle
(C) a right triangle
(D) an isosceles triangle

Given: The ratio of angles of triangles is 5 : 3: 7

Let angles of the triangle be $\mathrm{\angle }A,\mathrm{\angle }B$ $and$ $\mathrm{\angle }C$

$\mathrm{\angle }A:\mathrm{\angle }B:\mathrm{\angle }C$

$5:3:7$$5x:3x:7x$$\text{ here x = constant }$

Then, $\mathrm{\angle }A=5x$

$\mathrm{\angle }B=3x$

$\mathrm{\angle }C=7x$

In $\mathrm{△}ABC$

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

$5x+3x+7x={180}^{\circ }$

$15x={180}^{\circ }$

$x=\frac{180}{18}={12}^{\circ }$

$\mathrm{\angle }A=5x=5\times {12}^{\circ }={60}^{\circ }$

$\mathrm{\angle }B=3x=3\times {12}^{\circ }={36}^{\circ }$

$\mathrm{\angle }C=7x=7\times {12}^{\circ }={84}^{\circ }$

All angles are less than 90° hence acute.

Question:5 Write the correct answer in each of the following:
If one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles can be

(A) $5^{\circ }$

(B) ${65}^{\circ }$

(C) ${145}^{\circ }$

(D) ${155}^{\circ }$

In $\mathrm{△}ABC$

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

[Sum of all interior angles of triangle is ${180}^{\circ }$]

$\Rightarrow \frac{1}{2}\mathrm{\angle }A+\frac{1}{2}\mathrm{\angle }B+\frac{1}{2}\mathrm{\angle }C=$ $\frac{{180}^{\circ }}{2}$ $={90}^{\circ }$

$\Rightarrow \frac{1}{2}\mathrm{\angle }B+\frac{1}{2}\mathrm{\angle }C={90}^{\circ }-\frac{1}{2}\mathrm{\angle }A................(1)$

$\text{Since, in}\mathrm{△}BOC$

$\text{As BO and OC are the angle bisectors of }\mathrm{\angle }ABC$$\text{and }\mathrm{\angle }BCA$,

$\frac{\mathrm{\angle }B}{2}+\frac{\mathrm{\angle }C}{2}+\mathrm{\angle }BOC={180}^{\circ }..................(2)$

Put value of equation (1) in equation (2)

${90}^{\circ }-\frac{1}{2}\mathrm{\angle }A+\mathrm{\angle }BOC={180}^{\circ }$
$\mathrm{\angle }BOC={180}^{\circ }-{90}^{\circ }+\frac{1}{2}\mathrm{\angle }A$

$\mathrm{\angle }BOC={90}^{\circ }+\frac{1}{2}\mathrm{\angle }A$

$\therefore \mathrm{\angle }A={130}^{\circ }$

$\mathrm{\angle }BOC={90}^{\circ }+\frac{1}{2}\times {130}^{\circ }$

$\mathrm{\angle }BOC={90}^{\circ }+{65}^{\circ }$

$\mathrm{\angle }BOC={155}^{\circ }$

Question:6 Write the correct answer in each of the following:
In Fig. 6.2, POQ is a line. The value of x is


$(A)$ ${20}^{\circ }$
$(B)$ ${25}^{\circ }$
$(C)$ ${30}^{\circ }$
$(D)$ ${35}^{\circ }$

Answer:

$\mathrm{\angle }POQ={180}^{\circ }$(angle on a straight line)

${40}^{\circ }+4x+3x={180}^{\circ }$

$7x={140}^{\circ }$

$x={20}^{\circ }$

Question:7 Write the correct answer in each of the following:
$\text{ In Fig. 6.3, if }OP\parallel RS$, $\mathrm{\angle }OPQ={110}^{\circ }\text{and }$ $\mathrm{\angle }QRS={130}^{\circ }\text{, then }$$\mathrm{\angle }PQR\text{ is equal to }$

$(A){40}^{\circ }$
$(B)$ ${50}^{\circ }$
$(C)$ ${60}^{\circ }$
$(D)$ ${70}^{\circ }$

Answer:

$\mathrm{\angle }RWV+\mathrm{\angle }SRW={180}^{\circ }$ $\text{ (sum of co- interior angles) }$ ${130}^{\circ }+\mathrm{\angle }RWV={180}^{\circ }$

$\mathrm{\angle }RWV={50}^{\circ }$
$\mathrm{\angle }PWQ=\mathrm{\angle }RWV={50}^{\circ }\text{(vertically opposite angles are equal) }$

Also, for line OP

$\mathrm{\angle }OPQ+\mathrm{\angle }WPQ={180}^{\circ }\text{(linear pair)}$

$\mathrm{\angle }WPQ={180}^{\circ }-\mathrm{\angle }OPQ$

$={180}^{\circ }-{110}^{\circ }$

$\mathrm{\angle }WPQ={70}^{\circ }$

Now, we know that the sum of the angles of a triangle is 180⁰

$\text{in }\mathrm{△}PQW$,

$\mathrm{\angle }WPQ+\mathrm{\angle }PQW+\mathrm{\angle }PWQ={180}^{\circ }$

$\mathrm{\angle }PQW={180}^{\circ }-\mathrm{\angle }WPQ-\mathrm{\angle }PWQ={180}^{\circ }-{70}^{\circ }-{50}^{\circ }$

$\mathrm{\angle }PQR={180}^{\circ }-{120}^{\circ }$

$\mathrm{\angle }PQR={60}^{\circ }$

Hence, option C is the correct answer

Question:8 Write the correct answer in each of the following: Angles of a triangle are in the ratio 2: 4 : 3. The smallest angle of the triangle is
$(A)$ ${60}^{\circ }$
$(B)$ ${40}^{\circ }$
$(C)$ ${80}^{\circ }$
$(D)$ ${20}^{\circ }$

Answer:

Hence, option (B) is correct

Question:1 For what value of x+y in Fig. 6.4 will ABC b,e a line? Justify our answer.

Answer:

Justification: As we know if the sum of two adjacent angles is 180⁰ it represents the straight line.

$\mathrm{\angle }CBD=x,$ $\mathrm{\angle }ABD=y$

$\text{To form ABC; a straight line the sum of}$$\mathrm{\angle }ABD+\mathrm{\angle }CBD={180}^{\circ }$

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

Question:2 Can a triangle have all angles less than 600? Give a reason for your answer.

Answer:

So it is not possible to have all angles less than 60⁰ because their sum will not be equal to 180⁰

Hence, the answer is No

Question:3 Can a triangle have two obtuse angles? Give a reason for your answer

No, a triangle91 °nnot have two obtuse angles
Reason: the obtuse angle is greater than 90⁰
Let us consider the smallest obtuse angle, i.e., 91⁰
Let the two obtuse angles be equal to 91^o, so the sum of these two obtuse angles is equal to 182⁰
But the sum of the interior angles of a triangle is always equal to 180⁰
So the given condition is not possible.

Hence, we cannot create a triangle with two obtuse angles

Question:4: How many triangles can be drawn having angles of 450, 640, and 720? Give a reason for your answer.

The three angles given are 45⁰, 64⁰ and 72⁰

Now, the sum of these angles = 45⁰+64⁰+72⁰

= 181^o

The sum of angles of a triangle cannot be more than 180⁰.

Hence, the triangle is not possible
So the answer is zero

Question:5: How many triangles can be drawn having angles 530, 640, and 630? Give a reason for your answer.

Answer:

Infinitely many triangles can be drawn having angles as 53⁰, 64^0, and 63⁰

The sum of given angles = 53⁰ + 64⁰⁺ 63⁰ = 180⁰

Here, we can see that the sum of all interior angles of triangles is 180⁰, so infinitely many triangles can be drawn depending on the lengths of their sides.

So, infinitely many triangles can be drawn from the given angles.

Question:6 In Fig. 6.5, find the value of x for which the lines l and m are parallel.

In the given °figure, the lines l and m are parallel.

Line n is a transversal line.

So, $\mathrm{\angle }x$and 44^o are co-interior interior angles.

We know that the sum of co-interior angles is 180^o.

Hencegle x + ${44}^{\circ }={180}^{\circ }$

$\mathrm{\angle }x={136}^{\circ }$ is the correct answer

Question:7 Each of these angles doesn't need tot each of these angles will be a right angle. Justify your answer.

Given that two adjacent angles are equal.

No, it is not necessary that each of these angles will be a right angle.
Let us see why.
Adjacent angles: Two angles are said to be adjacent only when they have a common vertex, a common side, but they do not overlap.
The following figure shows adjacent angles ($\mathrm{\angle }1$ and$\mathrm{\angle }2$ ), ($\mathrm{\angle }DBC,$\angle CBA$)

But in the case that bot:h adjacent angles are right angles, they should form a linear pair, or we can say that they should lie on the same line, but such a condition is not given to us.

Question:8 If one of the angles formed by two intersecting lines is a right angle, what can you say about the other three angles? Give a reason for your answer.

Answer:

Let AB and CD be two intersecting lines that intersect at point O.

If one of the angles is formed by two intersecting lines is a right angle,
$\text{Then let}$$\mathrm{\angle }AOD={90}^{\circ }$
$\mathrm{\angle }AOD=\mathrm{\angle }BOC={90}^{\circ }$$\text{(vertically opposite angles)}$
$\mathrm{\angle }AOD,\mathrm{\angle }BOD$$\text{form lin,ear pair}$

$\mathrm{\angle }AOD+\mathrm{\angle }BOD={180}^{\circ }$
$\text{Hence,}$ $\mathrm{\angle }BOD={90}^{\circ }$
$\mathrm{\angle }BOD=\mathrm{\angle }AOC={90}^{\circ }$$\text{ (vertically opposite angles)}$

Hence all the other angles should be at right angles.

Question:9 In Fig.6.6, which of the two lines are parallel and why?

Answer:
Consider l and m, now n is the transversal
${132}^{\circ }+{48}^{\circ }={180}^{\circ }$ $(sumofco-interiorangles={180}^{\circ })$
Hence, the lines are parallel
Consider p and q, now r is the transversal

${106}^{\circ }+{73}^{\circ }={179}^{\circ }\ne {180}^{\circ }($ sum of co - interior angles $\ne {180}^{\circ })$
Hence, the lines are not parallel
Therefore, l and m are parallel

Question:10 Two lines l and m are perpendicular to the same line n. Are l and m perpendicular to each other? Give a reason for your answer.

The two lines l and m are perpendicular to the same line n.
We can have 3 possible scenarios:
1. The lines l and m may lie on the same side of n

$\mathrm{\angle }lAn=\mathrm{\angle }mCn={90}^{\circ }$
As we can see corresponding angles are equal, so l is parallel to m
l and m are not pe, perpendicular to each other
2. The lines l and m may lie on the opposite side of n
Extend lA to point B and mC to point D

$\mathrm{\angle }lAC=\mathrm{\angle }mCA={90}^{\circ }$
As we can see the interior opposite angles are equal, so l is parallel to m
l and m are not perpendicular to each other
3. The lines l and m are the same and overlap each other.
Again, l and m are not perpendicular to each other

Hence, the lines l and m are not perpendicular.

Question:1 In Fig. 6.9, OD is the bisector of, OE is the bisector of, and show that the points A, O, and B are collinear.

Answer:

Given in the figure $OD\perp OE.$, OD and OE are bisectors of $\mathrm{\angle }AOC$ and $\mathrm{\angle }BOC$ respectively
To show: Points A, O & B are collinear, i.e., AOB is a straight line
Proof: Since O, D, and OE bisect angles $\mathrm{\angle }AOC$ and $\mathrm{\angle }BOC$ respectively
$\mathrm{\angle }AOC=2\mathrm{\angle }DOC$… (i)
And $\mathrm{\angle }COB=2\mathrm{\angle }COE$ … (ii)
On adding equations (i) and (ii) we get
$\mathrm{\angle }AOC+\mathrm{\angle }COB=2\mathrm{\angle }DOC+2\mathrm{\angle }COE$
$\mathrm{\angle }AOC+\mathrm{\angle }COB=2(\mathrm{\angle }DOC+\mathrm{\angle }COE)$
$\mathrm{\angle }AOC+\mathrm{\angle }COB=2\mathrm{\angle }DOE$
$\mathrm{\angle }AOC+\mathrm{\angle }COB=2\times {90}^{\circ }$ (Given $OD\perp OE$)
$\mathrm{\angle }AOC+\mathrm{\angle }COB={180}^{\circ }$
$\mathrm{\angle }AOB={180}^{\circ }$
So, $\mathrm{\angle }AOC$ & $\mathrm{\angle }COB$ are forming a linear pair.
AOB is a straight line.
Therefore, points A and B are collinear.
Hence proved

Question:2 $\text{In Fig. 6.10,}\mathrm{\angle }1={60}^{\circ }$ $\text{ and}$ $\mathrm{\angle }6={120}^{\circ }$. Show that the lines m and n are parallel.

Answer:

Given: The figure $\mathrm{\angle }1={60}^{\circ }$ and $\mathrm{\angle }6={120}^{\circ }$
To show : $m\parallel n$
Proof: $\mathrm{\angle }1={60}^{\circ }$ and $\mathrm{\angle }6={120}^{\circ }$
Here, $\mathrm{\angle }1={60}^{\circ }$ and $\mathrm{\angle }6={120}^{\circ }$
Here $\mathrm{\angle }1=\mathrm{\angle }3$ (Vertically opposite angles)
$\mathrm{\angle }3=\mathrm{\angle }1={60}^{\circ }$
Now, $\mathrm{\angle }3+\mathrm{\angle }6={60}^{\circ }+{120}^{\circ }$
$\mathrm{\angle }3+\mathrm{\angle }6={180}^{\circ }$
We know if the sum of two interior angles on the same side is ${180}^{\circ }$, then the lines are parallel.el.
Hence proved, $m\parallel n$.

Question:3 AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (Fig. 6.11). Show that AP is parallel to BQ.

Answer:

Given: I,n the figure, $l\parallel m$, AP and BQ are the Bisectors of $\mathrm{\angle }EAB$ and $\mathrm{\angle }ABH$
To prove: $AP\parallel BQ$
Proof: Since $l\parallel m$ and t is a transversal therefore
$\mathrm{\angle }EAB=\mathrm{\angle }ABH$ (alternate interior angles)

$\frac{1}{2}\mathrm{\angle }EAB=\frac{1}{2}\mathrm{\angle }ABH$(Divide both sides by 2)
$\mathrm{\angle }PAB=\mathrm{\angle }ABQ$ (AP & BQ are the bisectors of $\mathrm{\angle }EAB$ & $\mathrm{\angle }ABH$)
Now consider, two lines AP and BQ with transversal AB
$\mathrm{\angle }PAB$and $\mathrm{\angle }ABQ$ are alternate interior angles, and these are equal.
Hence, $AP\parallel BQ$
Hence proved

Question:4. If in Figure bisectors AP and BQ of the alternate interior angles are parallel, then show that.

Answer:

In the figure $AP\parallel BQ$, AP and BQ are the bisectors of the alternate interior angles $\mathrm{\angle }CAB$ and $\mathrm{\angle }ABF$.

To show : $l\Vert m$
Proof: Since $AP\Vert BQ$ and $t$ is transversal therefore
$\mathrm{\angle }PAB=\mathrm{\angle }ABQ$ (Alternate Interior Angles)
$\Rightarrow 2\mathrm{\angle }PAB=2\mathrm{\angle }ABQ$ (Multiplying both sides by 2 )
Now, AP and BQ are the bisectors of alternate interior angles $\mathrm{\angle }CAB$ and $\mathrm{\angle }ABF$
$2\mathrm{\angle }PAB=\mathrm{\angle }CAB$
$2\mathrm{\angle }ABQ=\mathrm{\angle }ABF$
So,
$\mathrm{\angle }CAB=\mathrm{\angle }ABF$
Now consider lines l and m
$\mathrm{\angle }CAB=\mathrm{\angle }ABF$ (alternate interior angles are equal)
Hence $l\parallel m$

Hence proved

Question:5 In Fig. 6.12, $BA\parallel EDandBC\parallel EF$. Show that $\mathrm{\angle }ABC=\mathrm{\angle }DEF$ [Hint: Produce DE to intersect BC at P (say)].

Answer:

Produce DE to intersect BC at P

Now, $EF\parallel BC$ and DP is the transversal
$\mathrm{\angle }DEF=\mathrm{\angle }DPC$ (corresponding angles) … (i)
Now $AB\parallel DP$ and BC is transversal,
$\mathrm{\angle }DPC=\mathrm{\angle }ABC$ (corresponding angles) … (ii)
From (i) and (ii) we get
$\mathrm{\angle }ABC=\mathrm{\angle }DEF$

Hence, Proved

Question:6 In Fig. 6.13, $BA\parallel ED$ and $BC\parallel EF$. Show that $\mathrm{\angle }ABC+\mathrm{\angle }DEF={180}^{\circ }$

Answer:

Given
$BA\parallel ED$ and $BC\parallel EF$
To show $\mathrm{\angle }ABC+\mathrm{\angle }DEF={180}^{\circ }$
Construction:
Extend EF to point P on AB

Proof: In figure, $BC\parallel EF$, so $BC\parallel PF$
$\because \mathrm{\angle }EPB+\mathrm{\angle }PBC={180}^{\circ }$ (Sum of co-interior angles is ${180}^{\circ }$) …(i)
Now, $AB\parallel ED$ and PE is the transversal line.
$\mathrm{\angle }EPB=\mathrm{\angle }DEF$ (Corresponding angle) …(ii)
From Equations (i) and (ii)
$\mathrm{\angle }DEF+\mathrm{\angle }PBC={180}^{\circ }$
$\mathrm{\angle }ABC+\mathrm{\angle }DEF={180}^{\circ }(\because \mathrm{\angle }PBC=\mathrm{\angle }ABC)$

Hence proved

Question:7. In Fig. 6.14, $DE\parallel QR$ and AP and BP are bisectors of $\mathrm{\angle }EAB$ and $\mathrm{\angle }RBA$, respectively. Find $\mathrm{\angle }APB.$

Answer: $\mathrm{\angle }APB={90}^{\circ }$

$DE\parallel QR$and AP and PB are the bisectors of $\mathrm{\angle }EAB$ and $\mathrm{\angle }RBA$,
We know that the interior angles on the same side of the transversal are supplementary
So,
$\mathrm{\angle }EAB+\mathrm{\angle }RBA={180}^{\circ }$
$\frac{1}{2}\mathrm{\angle }EAB+\frac{1}{2}\mathrm{\angle }RBA=\frac{1}{2}({180}^{\circ })$
$\frac{1}{2}\mathrm{\angle }EAB+\frac{1}{2}\mathrm{\angle }RBA={90}^{\circ }$ …(i)
AP and BP are the bisectors,s of $\mathrm{\angle }EAB$ and $\mathrm{\angle }RBA$ respectively.
$\mathrm{\angle }BAP=\frac{1}{2}\mathrm{\angle }EAB$…(ii)
$\mathrm{\angle }ABP=\frac{1}{2}\mathrm{\angle }RBA$ …(iii)
On adding equations (ii) and (iii) we get
$\mathrm{\angle }BAP+\mathrm{\angle }ABP=\frac{1}{2}\mathrm{\angle }EAB+\frac{1}{2}\mathrm{\angle }RBA$
From equation (i)
$\mathrm{\angle }BAP+\mathrm{\angle }ABP={90}^{\circ }$
In $\mathrm{△}APB$,
$\mathrm{\angle }BAP+\mathrm{\angle }ABP+\mathrm{\angle }APB={180}^{\circ }$
${90}^{\circ }+\mathrm{\angle }APB={180}^{\circ }$
$\mathrm{\angle }APB={180}^{\circ }-{90}^{\circ }={90}^{\circ }$

Hence $\mathrm{\angle }APB={90}^{\circ }$

Question:8

The angles of a triangle are in the ratio 2:: 4. Find the value of each angle. What type of triangle is it?

Answer:

Angles of the triangle are in the ratio- 2 : 3: 4
Let the angles are 2x, 3x, 4x then:
$2x+3x+4x={180}^{\circ }$ (angle sum property)
$9x={180}^{\circ }$
$x={20}^{\circ }$
Then the angles of the triangle are:
$2x={40}^{\circ }$
$3x={60}^{\circ }$
$4x={80}^{\circ }$
This triangle is scalene as all the angles are of different measures.

Question:9 A triangle ABC is right-angled at A. L is a point on BC such that $AL\perp BC$. Prove that $\mathrm{\angle }BAL=\mathrm{\angle }ACB$.

Answer:

$AL\perp BC$ $\mathrm{\angle }BAL=\mathrm{\angle }ACB$

Proof: In $\mathrm{△}ABC$ and $\mathrm{△}LAB$,
$\mathrm{\angle }BAC=\mathrm{\angle }ALB$ (each ${90}^{\circ }$) (i)
And $\mathrm{\angle }ABC=\mathrm{\angle }ABL$ (Common angle) (ii)
In $\mathrm{△}ABC$,
$\mathrm{\angle }BAC+\mathrm{\angle }ABC+\mathrm{\angle }ACB={180}^{\circ }$ (angle sum property)
$\mathrm{\angle }ABC+\mathrm{\angle }ACB={180}^{\circ }-{90}^{\circ }$ (from i)
$\mathrm{\angle }ACB={90}^{\circ }-\mathrm{\angle }ABC$
In $\mathrm{△}ABL$,
$\mathrm{\angle }BAL+\mathrm{\angle }ALB+\mathrm{\angle }ABL={180}^{\circ }$(angle sum property)
$\mathrm{\angle }BAL+\mathrm{\angle }ABL={180}^{\circ }-{90}^{\circ }$ (from i)
$\mathrm{\angle }BAL={90}^{\circ }-\mathrm{\angle }ABL$
$\mathrm{\angle }BAL={90}^{\circ }-\mathrm{\angle }ABC$ (from ii)
Hence,$\mathrm{\angle }ACB={90}^{\circ }-\mathrm{\angle }ABC=\mathrm{\angle }BAL$

Hence proved

Question:10: Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

Answer:

Given: Two lines m and n are parallel, and another two lines p and q are respectively perpendicular to m and n, i.e., $p\perp m,p\perp n,q\perp m,q\perp n$.
To prove
$p\parallel q$

Proof:

Since m||n and p are perpendicular to m and n.
$\mathrm{\angle }5=\mathrm{\angle }6=\mathrm{\angle }7=\mathrm{\angle }8=\mathrm{\angle }11=\mathrm{\angle }12={90}^{\circ }$
Similarly, q is perpendicular to m and n.
$\mathrm{\angle }1=\mathrm{\angle }2=\mathrm{\angle }3=\mathrm{\angle }4=\mathrm{\angle }9=\mathrm{\angle }10={90}^{\circ }$
Now for lines p and q, m is the transversal
$\mathrm{\angle }1=\mathrm{\angle }2=\mathrm{\angle }3=\mathrm{\angle }4=\mathrm{\angle }5=\mathrm{\angle }6=\mathrm{\angle }7=\mathrm{\angle }8={90}^{\circ }$
So we can see that all the conditions are fulfilled for the lines to be parallel, i.e., Corresponding angles are equal, the sum of interior angles is 180^o, and alternate angles are equal.

Hence, $p\parallel q$

Hence proved

Question:1 If two lines intersect, prove that the vertically opposite angles are equal.

Answer:

It is given that if two lines intersect, the vertically opposite angles are equal.
Proof:

Now let AB and CD be two lines intersecting at point O.

From the figure, we have two pairs of vertically opposite angles, namely: ,

(i) $\mathrm{\angle }AOC$ and $\mathrm{\angle }BOD$

(ii) $\mathrm{\angle }AOD$ and $\mathrm{\angle }BOC$

Now we have to prove that $\mathrm{\angle }AOC=\mathrm{\angle }BOD$

And $\mathrm{\angle }AOD=\mathrm{\angle }BOC$

$\Rightarrow$ Now ray OA stands on line CD

$\mathrm{\angle }AOC+\mathrm{\angle }AOD={180}^{\circ }$ … (i) (linear pair angles)

Similarly, can we write

$\mathrm{\angle }AOD+\mathrm{\angle }BOD={180}^{\circ }$ … (ii) (linear pair angles)

From equation (i) and (ii) comparing

$\mathrm{\angle }AOC+\mathrm{\angle }AOD=\mathrm{\angle }AOD+\mathrm{\angle }BOD$

$\Rightarrow \mathrm{\angle }AOC=\mathrm{\angle }BOD$

Similarly, we can prove that $\mathrm{\angle }AOD=\mathrm{\angle }BOC$

Hence Proved.

Question:2 Bisectors of interior $\mathrm{\angle }B$ and exterior $\mathrm{\angle }ACD$ of a $\mathrm{△}ABC$ intersect at the point T. Prove that $\mathrm{\angle }BTC=\frac{1}{2}\mathrm{\angle }BAC$.

Answer:

According to the question,
Bisectors of interior $\mathrm{\angle }B$ and exterior $\mathrm{\angle }ACD$ of a $\mathrm{△}ABC$ intersect at the point T
$\mathrm{\angle }TBC=\frac{1}{2}\mathrm{\angle }ABC$… (1)
And $\mathrm{\angle }TCD=\frac{1}{2}\mathrm{\angle }ACD$ … (2)

Now from $\mathrm{△}ABC$ we have
$\mathrm{\angle }BAC+\mathrm{\angle }ABC=\mathrm{\angle }ACD$… (3) (exterior angle is equal to the sum of interior opposite angles)
And from $\mathrm{△}TBC$ we have
$\mathrm{\angle }BTC+\mathrm{\angle }TBC=\mathrm{\angle }TCD$ (exterior angle is equal to the sum of interior opposite angles)
Or $\mathrm{\angle }BTC+\frac{1}{2}\mathrm{\angle }ABC=\frac{1}{2}\mathrm{\angle }ACD$ using (1 and 2)
Or $\mathrm{\angle }BTC=\frac{1}{2}(\mathrm{\angle }ACD-\mathrm{\angle }ABC)$
Using (3), $\mathrm{\angle }ACD-\mathrm{\angle }ABC=\mathrm{\angle }BAC$
So,

$\mathrm{\angle }BTC=\frac{1}{2}\mathrm{\angle }BAC$

Hence Proved

Question:3 A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.

Answer:

Given: A transversal EF cuts two parallel lines AB and CD at points G & H. GL and HM are bisectors of angles

To prove: $GL\parallel HM$
Proof: $\mathrm{\angle }EGB=\mathrm{\angle }GHD$ (Corresponding angles)
$\frac{1}{2}\mathrm{\angle }EGB=\frac{1}{2}\mathrm{\angle }GHD$

$\mathrm{\angle }EGL=\mathrm{\angle }GHM$
These are the corresponding angles formed by the line GL and HM, where EF is the transversal.

$.]hereforeGL\parallel HM$

Hence proved

Question:4 Prove that through a given point, we can draw only one perpendicular to a given line. [Hint: Use proof by contradiction].

Answer:

Given: Consider a line R and a point P

Construction:
Draw two lines (m and n) passing through P, which are perpendicular to line R.
To prove: Only one perpendicular line can be drawn through a point P
Proof: In $\mathrm{△}APB$
$\mathrm{\angle }A+\mathrm{\angle }P+\mathrm{\angle }B={180}^{\circ }${angle sum property}

${90}^{\circ }+\mathrm{\angle }P+{90}^{\circ }={180}^{\circ }$
$\mathrm{\angle }P=180-{180}^{\circ }$
$\mathrm{\angle }P=0^{\circ }$
So lines n and m will coincide
Therefore, we can draw only one perpendicular to a given line.

Hence proved

Question:5 Prove that two lines that are respectively parallel to two intersecting lines intersect each other. [Hint: Use proof by contradiction].

Answer:

Given: Let lines be and y be two intersecting lines. Let n and p be another two lines which are perpendicular to x and y,

To prove: n and p interse,ct at a point
Proof: Let lines n and p are not intersecting then $n\parallel p$ … (1)
Since n and p are parallel and n is perpendicular to x, and p is perpendicular to y respectively

So, $x\parallel y$
But, it is a contradiction as it is given that x and y are two intersecting lines.
Thus our assumption is wrong.
n and p intersect at a point

Hence proved

Question:6 Prove that a triangle must have at least two acute angles.

Answer:

It is given that a triangle must have at least two acute angles.
An acute angle is less than 90 degrees
Let us assume that a triangle does not have two acute angles.
So, it has two angles that are either right angles (=90 degrees) or obtuse angles (greater than 90 degrees)
So let two right angles be present,
So, using the nanglesum property of a triangle, the third angle must be zero, which is not possible.
Also, let one angle be right and one be obtuse. We can take the smallest obtuse angle, i.e.,
So, using the angle sum property of a triangle, the third angle must be negative, which is not possible.
Again, if both the angles are obtuse, the third angle must be negative, which is not possible.
So a triangle must have at least two acute angles.

Hence proved

Question 7: In Fig. 6.17, $\mathrm{\angle }Q>\mathrm{\angle }R$, PA is the bisector of $\mathrm{\angle }QPR$ and $PM\perp QR$. Prove that $\mathrm{\angle }APM=\frac{1}{2}(\mathrm{\angle }Q-\mathrm{\angle }R)$.

Answer:

Given : In $\mathrm{△}PQR$, $\mathrm{\angle }Q>\mathrm{\angle }R$

PA is the bisector of $\mathrm{\angle }QPR$ and $PM\perp QR$.

To prove : $\mathrm{\angle }APM=\left(\frac{1}{2}\right)(\mathrm{\angle }Q-\mathrm{\angle }R)$
Proof: Since PA is the bisector of $\mathrm{\angle }P$, we have

$\mathrm{\angle }APQ=\left(\frac{1}{2}\right)\mathrm{\angle }P$ … (i)
In right angled $\mathrm{△}PMQ$ we have

$\mathrm{\angle }Q+\mathrm{\angle }MPQ+{90}^{\circ }={180}^{\circ }$(Angle sum property)

$\Rightarrow \mathrm{\angle }MPQ={90}^{\circ }-\mathrm{\angle }Q$ … (ii)
Now
$\mathrm{\angle }APM=\mathrm{\angle }APQ-\mathrm{\angle }MPQ$
$\mathrm{\angle }APM=\frac{1}{2}\mathrm{\angle }P-(90-\mathrm{\angle }Q)$ using (i) & (ii)
$\mathrm{\angle }APM=\frac{1}{2}\mathrm{\angle }P-{90}^{\circ }+\mathrm{\angle }Q$
$\mathrm{\angle }APM=\frac{1}{2}\mathrm{\angle }P-\frac{1}{2}(\mathrm{\angle }P+\mathrm{\angle }R+\mathrm{\angle }Q)+\mathrm{\angle }Q$Since ${90}^{\circ }=\frac{1}{2}(\mathrm{\angle }P+\mathrm{\angle }R+\mathrm{\angle }Q)$
$\mathrm{\angle }APM=\frac{1}{2}\mathrm{\angle }P-\frac{1}{2}\mathrm{\angle }P-\frac{1}{2}\mathrm{\angle }R-\frac{1}{2}\mathrm{\angle }Q+\mathrm{\angle }Q$
$\mathrm{\angle }APM=\frac{1}{2}(\mathrm{\angle }Q-\mathrm{\angle }R)$

Hence proved