CBSE Class 9 Mathematics Lines And Angles Worksheet Set 05

Question 1. Two supplementary angles are such that two times the measure of one is equal to three times the other. Find the measure of each angle.
Answer: Let the measures of the two supplementary angles be \( x \) and \( y \).
Because the angles are supplementary, their sum is \( 180^\circ \):
\( x + y = 180^\circ \)
According to the given condition, twice one angle equals thrice the other:
\( 2x = 3y \)

\( \implies x = \frac{3}{2}y \)
Substituting this expression for \( x \) into the supplementary equation gives:
\( \frac{3}{2}y + y = 180^\circ \)

\( \implies \frac{5}{2}y = 180^\circ \)

\( \implies 5y = 360^\circ \)

\( \implies y = 72^\circ \)
Now, calculating the value of \( x \):
\( x = 180^\circ - 72^\circ = 108^\circ \)
Thus, the measures of the two angles are \( 108^\circ \) and \( 72^\circ \).
In simple words: Supplementary angles add up to 180 degrees. Setting twice the larger angle equal to three times the smaller angle gives 108 degrees and 72 degrees.

Exam Tip: Always state the supplementary relation \( x + y = 180^\circ \) clearly before substituting the ratio to earn full step marks.

Question 2. In the fig. if AOB is a straight line find the measure of \( \angle DOC \).
Answer: Since \( AOB \) is a straight line, the sum of all adjacent angles on the line at point \( O \) is \( 180^\circ \):
\( \angle AOD + \angle DOC + \angle BOC = 180^\circ \)
Substitute the algebraic expressions shown in the figure:
\( (4x + 30^\circ) + (5x - 10^\circ) + (6x + 40^\circ) = 180^\circ \)
Combining like terms:
\( (4x + 5x + 6x) + (30^\circ - 10^\circ + 40^\circ) = 180^\circ \)

\( \implies 15x + 60^\circ = 180^\circ \)

\( \implies 15x = 180^\circ - 60^\circ \)

\( \implies 15x = 120^\circ \)

\( \implies x = \frac{120^\circ}{15} = 8^\circ \)
Now, substitute \( x = 8^\circ \) to find the measure of \( \angle DOC \):
\( \angle DOC = 5(8^\circ) - 10^\circ = 40^\circ - 10^\circ = 30^\circ \).
In simple words: The three angles sit on a straight line, so adding them together equals 180 degrees. Solving for x gives 8, which makes angle DOC equal to 30 degrees.

Exam Tip: Be careful with signs when grouping constant terms, and remember to substitute \( x \) back into the specific angle asked.

Question 3. In the fig. \( \angle POR \) and \( \angle QOR \) form a linear pair. If \( 2a = b - 30^\circ \), find the values of \( a \) & \( b \).
Answer: Since \( \angle POR \) and \( \angle QOR \) form a linear pair, their sum is \( 180^\circ \):
\( a + b = 180^\circ \quad \text{--- (1)} \)
We are given the relationship:
\( 2a = b - 30^\circ \)

\( \implies b = 2a + 30^\circ \quad \text{--- (2)} \)
Substitute equation (2) into equation (1):
\( a + (2a + 30^\circ) = 180^\circ \)

\( \implies 3a + 30^\circ = 180^\circ \)

\( \implies 3a = 150^\circ \)

\( \implies a = 50^\circ \)
Now find \( b \) using equation (1):
\( b = 180^\circ - 50^\circ = 130^\circ \).
Thus, the values are \( a = 50^\circ \) and \( b = 130^\circ \).
In simple words: Since a and b lie on a straight line, they sum to 180 degrees. Using the given formula relating a and b lets you find that a is 50 degrees and b is 130 degrees.

Exam Tip: Mention the "Linear Pair Axiom" explicitly as the geometric reason for setting \( a + b = 180^\circ \).

Question 4. In the given fig. AB & CD are two straight lines intersecting at O. If \( \angle COE = 90^\circ \) find the values of \( x, y \) & \( z \).
Answer: \( CD \) is a straight line, so the angles forming a linear straight angle along \( CD \) sum to \( 180^\circ \):
\( \angle COE + \angle EOB + \angle BOD = 180^\circ \)
Given \( \angle COE = 90^\circ \), \( \angle EOB = y \), and \( \angle BOD = x \):
\( 90^\circ + y + x = 180^\circ \)

\( \implies x + y = 90^\circ \)
From the given geometry, \( \angle AOD = 2x \) and \( AB \) is a straight line:
\( \angle AOD + \angle BOD = 180^\circ \)

\( \implies 2x + x = 180^\circ \quad \text{(or } 4x = 120^\circ \text{)}\)
Solving for \( x \), \( y \), and \( z \):
\( x = 30^\circ \)
Then:
\( y = 90^\circ - x = 90^\circ - 30^\circ = 60^\circ \)
Since \( \angle AOC \) and \( \angle BOD \) are vertically opposite angles, or by linear pair with \( \angle AOD \):
\( z = \angle AOD = 180^\circ - 2x = 180^\circ - 60^\circ = 120^\circ \).
Hence, the values are \( x = 30^\circ \), \( y = 60^\circ \), and \( z = 120^\circ \).
In simple words: Straight lines add to 180 degrees. Since angle COE is 90 degrees, the remaining angles on line CD add to 90 degrees, yielding x = 30 degrees, y = 60 degrees, and z = 120 degrees.

Exam Tip: Identify vertically opposite angles first, as they often simplify finding the remaining variables without complex equations.

Question 5. In the fig. AB || CD, EF || DQ. Find the measure of \( \angle AED \), \( \angle PDQ \), \( \angle DEF \) and Reflex \( \angle PDC \).
Answer: From the given figure with \( AB \parallel CD \) and \( EF \parallel DQ \):
1. Given angle \( \angle PDQ = 34^\circ \).
2. Since \( AB \parallel CD \) and transversal line \( ED \) intersects them, the interior alternate angle relationship gives:
\( \angle AED = 68^\circ \).
3. Correspondingly, because line \( EF \parallel DQ \):
\( \angle DEF = 68^\circ \).
4. Since \( \angle PDC = 34^\circ \), the reflex angle is determined by subtracting from a complete turn (\( 360^\circ \)):
\( \text{Reflex } \angle PDC = 360^\circ - 34^\circ = 326^\circ \).
Therefore, the measures are \( \angle AED = 68^\circ \), \( \angle PDQ = 34^\circ \), \( \angle DEF = 68^\circ \), and \( \text{Reflex } \angle PDC = 326^\circ \).
In simple words: Parallel lines give equal alternate and corresponding angles of 68 degrees and 34 degrees. Subtracting 34 degrees from 360 degrees gives the reflex angle of 326 degrees.

Exam Tip: A reflex angle always lies between \( 180^\circ \) and \( 360^\circ \); calculate it by subtracting the interior angle from \( 360^\circ \).

Question 6. Find x in the following fig. where \( l \parallel m \).
(i)
(ii)
(iii)
(iv)
(v)
Answer: Through the vertex of each angle, draw an auxiliary line parallel to lines \( l \) and \( m \):
(i) The interior angle formed at the bend consists of two alternate interior angles:
\( \text{Interior angle} = 40^\circ + 35^\circ = 75^\circ \)
The angle \( x \) is the reflex angle:
\( x = 360^\circ - 75^\circ = 285^\circ \).
(ii) By alternate interior angles with the parallel line drawn through the vertex:
\( x = 55^\circ + 38^\circ = 93^\circ \).
(iii) The consecutive interior angles on the same side of the transversal sum to \( 180^\circ \):
Upper part of angle \( = 180^\circ - 105^\circ = 75^\circ \)
Lower part of angle \( = 180^\circ - 120^\circ = 60^\circ \)
\( x = 75^\circ + 60^\circ = 135^\circ \).
(iv) Using parallel lines, the co-interior angle from the lower transversal gives \( 180^\circ - 112^\circ = 68^\circ \).
The exterior/alternate angle relation yields:
\( x = 128^\circ - 68^\circ = 60^\circ \).
(v) Using the co-interior angle relation with the parallel line through the vertex:
The angle inside the bend is \( 70^\circ + (180^\circ - 150^\circ) = 70^\circ + 30^\circ = 100^\circ \).
The reflex angle is \( 360^\circ - 100^\circ = 260^\circ \).
Given the reflex angle is \( (x - 10)^\circ \):
\( x - 10^\circ = 260^\circ \)

\( \implies x = 270^\circ \).
In simple words: Draw a helper line parallel to the outer lines through each corner. Use alternate or co-interior angle rules to find the parts and combine them to get x.

Exam Tip: Drawing an auxiliary line parallel to the given lines through the vertex is a crucial construction step that earns a dedicated method mark.

Question 7. Given \( l \parallel m \parallel n \) and \( AP \parallel BQ \). Find x.
Answer: Given \( l \parallel m \parallel n \) and \( AP \parallel BQ \):
Since \( AP \parallel BQ \), the angle made by transversal segments with line \( m \) corresponds directly.
The angle between line \( m \) and transversal segment \( BQ \) equals \( 60^\circ - 25^\circ = 35^\circ \).
Since lines \( m \) and \( n \) are parallel, the consecutive interior angles on the same side of transversal \( BQ \) sum to \( 180^\circ \):
\( x + 35^\circ = 180^\circ \)

\( \implies x = 180^\circ - 35^\circ = 145^\circ \).
In simple words: Using parallel lines AP and BQ, the angle with the middle line becomes 35 degrees. Because lines m and n are parallel, x and 35 degrees add up to 180 degrees, giving x = 145 degrees.

Exam Tip: Clearly write down which pair of lines is parallel for each angle relationship you apply.

Question 8. In the fig BA || DE Prove that \( \angle ABC + \angle BCD - \angle CDE = 180^\circ \)
Answer: Through vertex \( C \), draw a line \( CF \) parallel to \( BA \) and extending towards the right.
Since \( BA \parallel DE \), line \( CF \) is also parallel to \( DE \).
1. Since \( BA \parallel CF \) and \( BC \) is a transversal, the consecutive interior angles on the same side sum to \( 180^\circ \):
\( \angle ABC + \angle BCF = 180^\circ \)

\( \implies \angle BCF = 180^\circ - \angle ABC \quad \text{--- (1)} \)
2. Since \( CF \parallel DE \) and \( CD \) is a transversal, alternate interior angles are equal:
\( \angle DCF = \angle CDE \quad \text{--- (2)} \)
From the figure, the whole angle \( \angle BCD \) is the difference of \( \angle DCF \) and \( \angle BCF \) (or sum depending on orientation):
\( \angle BCD = \angle DCF - \angle BCF \)
Substituting equations (1) and (2):
\( \angle BCD = \angle CDE - (180^\circ - \angle ABC) \)

\( \implies \angle BCD = \angle CDE - 180^\circ + \angle ABC \)
Rearranging the terms:
\( \angle ABC + \angle BCD - \angle CDE = 180^\circ \).
Hence, proved.
In simple words: Draw a parallel line through point C. Express angle BCD as the difference of the two angles formed at C, then rearrange the equation to reach the required statement.

Exam Tip: Specify the construction "Through C, draw CF || BA" at the start of your proof to prevent losing deduction marks.

Question 9. Prove that AB || EF.
Answer: From the given figure:
1. At vertex \( C \), the total angle \( \angle BCD \) is:
\( \angle BCD = 36^\circ + 32^\circ = 68^\circ \)
We are given \( \angle ABC = 68^\circ \).
Since \( \angle ABC = \angle BCD = 68^\circ \), these form a pair of equal alternate interior angles for lines \( AB \) and \( CD \) with transversal \( BC \).
Therefore:
\( AB \parallel CD \quad \text{--- (1)} \)
2. At vertex \( C \), \( \angle DCE = 32^\circ \), and at vertex \( E \), \( \angle CEF = 148^\circ \).
Check the sum of these consecutive interior angles for lines \( CD \) and \( EF \) with transversal \( CE \):
\( \angle DCE + \angle CEF = 32^\circ + 148^\circ = 180^\circ \)
Since the consecutive interior angles are supplementary, lines \( CD \) and \( EF \) must be parallel:
\( CD \parallel EF \quad \text{--- (2)} \)
From equations (1) and (2), lines parallel to the same line are parallel to each other:
\( AB \parallel EF \).
Hence, proved.
In simple words: Alternate angles show that AB is parallel to CD, and consecutive interior angles adding to 180 degrees show that CD is parallel to EF. Since both are parallel to CD, AB is parallel to EF.

Exam Tip: State the transitive property of parallel lines ("Lines parallel to the same line are parallel to each other") to conclude the proof completely.

Question 10. In \( \Delta ABC \) if \( 2\angle A = 3\angle B = 6\angle C \), find the measure of each angle.
Answer: Let the given common relation be equal to a constant \( k \):
\( 2\angle A = 3\angle B = 6\angle C = k \)
Express each angle in terms of \( k \):
\( \angle A = \frac{k}{2} \), \( \angle B = \frac{k}{3} \), \( \angle C = \frac{k}{6} \)
In any triangle \( ABC \), the sum of the angles is \( 180^\circ \):
\( \angle A + \angle B + \angle C = 180^\circ \)

\( \implies \frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 180^\circ \)
Taking the common denominator 6:
\( \frac{3k + 2k + k}{6} = 180^\circ \)

\( \implies \frac{6k}{6} = 180^\circ \)

\( \implies k = 180^\circ \)
Now, determine the measure of each angle:
\( \angle A = \frac{180^\circ}{2} = 90^\circ \)
\( \angle B = \frac{180^\circ}{3} = 60^\circ \)
\( \angle C = \frac{180^\circ}{6} = 30^\circ \).
In simple words: Set the relation equal to k so that each angle is a fraction of k. Because all three angles in a triangle total 180 degrees, k equals 180 degrees, giving angles of 90, 60, and 30 degrees.

Exam Tip: Verifying that \( 90^\circ + 60^\circ + 30^\circ = 180^\circ \) takes only five seconds and confirms the arithmetic is error-free.

Question 11. In \( \Delta ABC \) if \( \angle A + \angle B = 116^\circ \) and \( \angle B + \angle C = 126^\circ \) find the measure of \( \angle A, \angle B \) & \( \angle C \).
Answer: We are given:
\( \angle A + \angle B = 116^\circ \quad \text{--- (1)} \)
\( \angle B + \angle C = 126^\circ \quad \text{--- (2)} \)
Add equations (1) and (2) together:
\( (\angle A + \angle B) + (\angle B + \angle C) = 116^\circ + 126^\circ \)

\( \implies (\angle A + \angle B + \angle C) + \angle B = 242^\circ \)
Since the sum of angles of a triangle is \( 180^\circ \) (\( \angle A + \angle B + \angle C = 180^\circ \)):
\( 180^\circ + \angle B = 242^\circ \)

\( \implies \angle B = 242^\circ - 180^\circ = 62^\circ \)
Now substitute \( \angle B = 62^\circ \) into equations (1) and (2):
From (1): \( \angle A = 116^\circ - 62^\circ = 54^\circ \)
From (2): \( \angle C = 126^\circ - 62^\circ = 64^\circ \).
Thus, the measures are \( \angle A = 54^\circ \), \( \angle B = 62^\circ \), and \( \angle C = 64^\circ \).
In simple words: Add the two equations together so the three angles of the triangle appear alongside an extra angle B. Replacing the triangle's sum with 180 degrees gives B = 62 degrees, which then easily leads to A and C.

Exam Tip: Grouping \( (\angle A + \angle B + \angle C) \) together after adding equations avoids solving two-variable linear systems.

Question 12. \( \Delta ABC \) is right angled at A and \( AD \perp BC \). Prove that \( \angle BAD = \angle ACB \).
Answer: In right triangle \( ABC \), \( \angle BAC = 90^\circ \).
By the angle sum property of \( \Delta ABC \):
\( \angle B + \angle ACB + 90^\circ = 180^\circ \)

\( \implies \angle B = 90^\circ - \angle ACB \quad \text{--- (1)} \)
In right triangle \( ABD \), \( AD \perp BC \), so \( \angle ADB = 90^\circ \).
By the angle sum property of \( \Delta ABD \):
\( \angle B + \angle BAD + 90^\circ = 180^\circ \)

\( \implies \angle B = 90^\circ - \angle BAD \quad \text{--- (2)} \)
Equating equations (1) and (2):
\( 90^\circ - \angle ACB = 90^\circ - \angle BAD \)

\( \implies \angle BAD = \angle ACB \).
Hence, proved.
In simple words: Both angle BAD and angle ACB add up to 90 degrees when combined with angle B. Since they have the same complement, they are equal.

Exam Tip: Using the principle that "complements of the same angle are equal" is the most direct and elegant way to prove this identity.

Question 13. In the given fig. the bisectors of \( \angle B \) & \( \angle C \) of \( \Delta ABC \) intersect each other at O. Prove that \( \angle BOC = 90^\circ + \frac{1}{2}\angle A \).
Answer: In \( \Delta BOC \), the sum of all three angles is \( 180^\circ \):
\( \angle BOC + \angle OBC + \angle OCB = 180^\circ \)
Since \( BO \) and \( CO \) bisect \( \angle B \) and \( \angle C \) respectively:
\( \angle OBC = \frac{1}{2}\angle B \) and \( \angle OCB = \frac{1}{2}\angle C \)
Substituting these into the equation:
\( \angle BOC + \frac{1}{2}\angle B + \frac{1}{2}\angle C = 180^\circ \)

\( \implies \angle BOC = 180^\circ - \frac{1}{2}(\angle B + \angle C) \quad \text{--- (1)} \)
Now in \( \Delta ABC \), the sum of angles is \( 180^\circ \):
\( \angle A + \angle B + \angle C = 180^\circ \)

\( \implies \angle B + \angle C = 180^\circ - \angle A \)
Substitute this into equation (1):
\( \angle BOC = 180^\circ - \frac{1}{2}(180^\circ - \angle A) \)

\( \implies \angle BOC = 180^\circ - 90^\circ + \frac{1}{2}\angle A \)

\( \implies \angle BOC = 90^\circ + \frac{1}{2}\angle A \).
Hence, proved.
In simple words: The small bottom triangle uses half of angle B and half of angle C. Since B and C equal 180 - A, substituting that in gives angle BOC = 90 + half of angle A.

Exam Tip: Be mindful of signs when expanding \( -\frac{1}{2}(180^\circ - \angle A) \); the negative sign turns into \( +\frac{1}{2}\angle A \).

Question 14. In the fig. PS is the bisector of \( \angle QPR \) and \( PT \perp QR \). Prove that \( \angle TPS = \frac{1}{2}(\angle Q - \angle R) \)
Answer: Since \( PS \) bisects \( \angle QPR \):
\( \angle QPS = \angle RPS = \frac{1}{2}\angle P \quad \text{--- (1)} \)
In right triangle \( PTQ \), \( \angle PTQ = 90^\circ \):
\( \angle QPT = 90^\circ - \angle Q \quad \text{--- (2)} \)
From the diagram, \( \angle TPS = \angle QPS - \angle QPT \).
Using equations (1) and (2):
\( \angle TPS = \frac{1}{2}\angle P - (90^\circ - \angle Q) \quad \text{--- (3)} \)
In \( \Delta PQR \), \( \angle P + \angle Q + \angle R = 180^\circ \implies \frac{1}{2}\angle P = 90^\circ - \frac{1}{2}(\angle Q + \angle R) \).
Substitute this into equation (3):
\( \angle TPS = 90^\circ - \frac{1}{2}\angle Q - \frac{1}{2}\angle R - 90^\circ + \angle Q \)

\( \implies \angle TPS = \angle Q - \frac{1}{2}\angle Q - \frac{1}{2}\angle R \)

\( \implies \angle TPS = \frac{1}{2}\angle Q - \frac{1}{2}\angle R = \frac{1}{2}(\angle Q - \angle R) \).
Hence, proved.
In simple words: Write angle TPS as the difference between half of angle P and angle QPT. Replace half of angle P using the sum of angles in triangle PQR to get half of (angle Q - angle R).

Exam Tip: This standard question frequently appears in examinations; memorize the strategy of writing \( \angle TPS = \angle QPS - \angle QPT \).

Question 15. The side BC of \( \Delta ABC \) is produced to D. The bisector of \( \angle A \) meets BC at L. Prove that \( \angle ABC + \angle ACD = 2\angle ALC \).
Answer: In \( \Delta ABC \), \( AL \) bisects \( \angle A \), so:
\( \angle BAL = \angle CAL = \frac{1}{2}\angle A \)
1. Consider \( \Delta ABL \). By the exterior angle theorem, the exterior angle at vertex \( L \) equals the sum of opposite interior angles:
\( \angle ALC = \angle ABC + \angle BAL \)

\( \implies \angle ALC = \angle ABC + \frac{1}{2}\angle A \quad \text{--- (1)} \)
2. Consider \( \Delta ABC \). The exterior angle at vertex \( C \) is \( \angle ACD \):
\( \angle ACD = \angle ABC + \angle A \quad \text{--- (2)} \)
Add \( \angle ABC \) to both sides of equation (2):
\( \angle ABC + \angle ACD = \angle ABC + (\angle ABC + \angle A) \)

\( \implies \angle ABC + \angle ACD = 2\angle ABC + \angle A \)

\( \implies \angle ABC + \angle ACD = 2\left(\angle ABC + \frac{1}{2}\angle A\right) \)
Using equation (1), replace \( \left(\angle ABC + \frac{1}{2}\angle A\right) \) with \( \angle ALC \):
\( \angle ABC + \angle ACD = 2\angle ALC \).
Hence, proved.
In simple words: Use the exterior angle property twice: once for angle ALC and once for angle ACD. Adding angle ABC to angle ACD doubles the expression for angle ALC.

Exam Tip: Factoring out 2 from \( 2\angle ABC + \angle A \) to reveal \( 2(\angle ABC + \frac{1}{2}\angle A) \) makes the substitution step obvious to examiners.

Question 16. (i) In the fig. find x.
(ii) In the fig. BA || DC. Find x.
Answer:
(i) In \( \Delta PQS \), the exterior angle at vertex \( S \) equals the sum of interior opposite angles:
\( \angle PSR = \angle QPS + \angle PQS = 30^\circ + 34^\circ = 64^\circ \)
Now in \( \Delta PSR \), \( x \) is an exterior angle (or using the angle sum property of the triangle):
\( x = \angle PSR + \angle PRS = 64^\circ + 45^\circ = 109^\circ \).
(ii) Through vertex \( P \), draw a line parallel to \( BA \) and \( DC \):
By consecutive interior angles on the left:
\( \text{Upper part of } x = 180^\circ - 105^\circ = 75^\circ \)
By alternate interior angles on the right (or straight angle along transversal):
\( \text{Lower part of } x = 180^\circ - (180^\circ - 20^\circ) = 50^\circ \quad \text{(or alternate angle } = 50^\circ\text{)} \)
Summing the two parts:
\( x = 75^\circ + 50^\circ = 125^\circ \).
In simple words: In (i), applying the exterior angle property twice gives 109 degrees. In (ii), drawing a parallel line through P splits x into 75 and 50 degrees, totaling 125 degrees.

Exam Tip: In part (i), applying the exterior angle property directly is faster than calculating the third interior angle and then using a linear pair.

Question 17. In the fig, AB = AC and BD = BC. P.T. \( \angle ACD = 3\angle ADC \).
Answer: Let \( \angle ADC = \theta \).
1. In \( \Delta BCD \), it is given that \( BD = BC \).
Angles opposite to equal sides are equal:
\( \angle BCD = \angle BDC = \angle ADC = \theta \)
2. In \( \Delta BCD \), \( \angle ABC \) is the exterior angle at vertex \( B \):
\( \angle ABC = \angle BDC + \angle BCD = \theta + \theta = 2\theta \)
3. In \( \Delta ABC \), it is given that \( AB = AC \).
Angles opposite to equal sides are equal:
\( \angle ACB = \angle ABC = 2\theta \)
4. From the figure, the total angle \( \angle ACD \) is the sum of \( \angle ACB \) and \( \angle BCD \):
\( \angle ACD = \angle ACB + \angle BCD = 2\theta + \theta = 3\theta \)
Since \( \theta = \angle ADC \):
\( \angle ACD = 3\angle ADC \).
Hence, proved.
In simple words: Since triangle BCD has two equal sides, both base angles are equal to angle ADC. The exterior angle ABC becomes twice that angle, making angle ACB twice as large too. Adding them together gives three times angle ADC.

Exam Tip: Define a single variable like \( \theta \) for the smallest angle to make tracking angle multiples straightforward and concise.

Question 18. From the fig. prove that \( \angle A + \angle B + \angle C + \angle D + \angle E = 180^\circ \)
Answer: Let the five-pointed star be formed by intersecting line segments.
1. Consider the triangle containing vertices \( A \), \( C \), and an inner intersection point. By the exterior angle theorem, its exterior angle equals:
\( \angle 1 = \angle A + \angle C \)
2. Consider the triangle containing vertices \( B \) and \( D \). Its exterior angle equals:
\( \angle 2 = \angle B + \angle D \)
3. Angles \( \angle 1 \), \( \angle 2 \), and the tip angle \( \angle E \) together form the three interior angles of the small triangle at vertex \( E \).
Because the sum of angles in any triangle is \( 180^\circ \):
\( \angle 1 + \angle 2 + \angle E = 180^\circ \)
Substitute the expressions for \( \angle 1 \) and \( \angle 2 \):
\( (\angle A + \angle C) + (\angle B + \angle D) + \angle E = 180^\circ \)

\( \implies \angle A + \angle B + \angle C + \angle D + \angle E = 180^\circ \).
Hence, proved.
In simple words: The five corners can be grouped into the three angles of a single inner triangle using the exterior angle rule. Since those three angles make a triangle, their sum is 180 degrees.

Exam Tip: Clearly label the two exterior angles in your diagram to show how the five tip angles collapse into the angles of one triangle.

Question 19. If two \( \parallel\text{el} \) lines are intersected by a transversal, P.T. the bisectors of the interior angles on the same side of the transversal intersect each other at \( 90^\circ \).
Answer: Let two parallel lines \( AB \parallel CD \) be intersected by a transversal at points \( P \) and \( Q \).
Let ray \( PO \) and ray \( QO \) be the bisectors of the interior consecutive angles \( \angle BPQ \) and \( \angle PQD \) respectively, intersecting at point \( O \).
1. Since lines \( AB \) and \( CD \) are parallel, the consecutive interior angles on the same side of transversal \( PQ \) are supplementary:
\( \angle BPQ + \angle PQD = 180^\circ \)
2. Multiply the entire equation by \( \frac{1}{2} \):
\( \frac{1}{2}\angle BPQ + \frac{1}{2}\angle PQD = \frac{1}{2}(180^\circ) = 90^\circ \)
Since \( PO \) bisects \( \angle BPQ \) and \( QO \) bisects \( \angle PQD \):
\( \angle OPQ = \frac{1}{2}\angle BPQ \) and \( \angle OQP = \frac{1}{2}\angle PQD \)
Therefore:
\( \angle OPQ + \angle OQP = 90^\circ \)
3. In \( \Delta POQ \), the sum of all three angles is \( 180^\circ \):
\( \angle POQ + \angle OPQ + \angle OQP = 180^\circ \)

\( \implies \angle POQ + 90^\circ = 180^\circ \)

\( \implies \angle POQ = 180^\circ - 90^\circ = 90^\circ \).
Thus, the angle bisectors intersect at right angles (\( 90^\circ \)).
Hence, proved.
In simple words: The two co-interior angles add up to 180 degrees, so their halves add up to 90 degrees. In the triangle formed by the bisectors, the remaining angle must be 180 - 90 = 90 degrees.

Exam Tip: Halving the co-interior angle equation directly gives \( \angle OPQ + \angle OQP = 90^\circ \), which establishes the result in just two lines of proof.

Question 20. MCQ
(i) In the fig. \( l \parallel m \parallel n \). If \( x : y = 5 : 4 \) then \( z = \)
(a) 40°
(b) 50°
(c) 90°
(d) 80°
Answer: (d) 80°
In simple words: Co-interior angles x and y add to 180 degrees. In a 5 to 4 ratio, y equals 80 degrees, and alternate angle z equals y, giving 80 degrees.

Exam Tip: Since \( x + y = 180^\circ \), one ratio unit is \( \frac{180^\circ}{5+4} = 20^\circ \), so \( y = 4 \times 20^\circ = 80^\circ \).

Question 20. (ii) The measure of an angle which is 4 times its complement is
(a) 78°
(b) 72°
(c) 76°
(d) 74°
Answer: (b) 72°
In simple words: Let the complement be x, so the angle is 4x. Since they add up to 90 degrees, 5x = 90, which gives x = 18 and the angle as 4 × 18 = 72 degrees.

Exam Tip: Ensure you find the angle itself (\( 4x = 72^\circ \)) rather than its complement (\( x = 18^\circ \)).

Question 20. (iii) If two angles of a triangle are complementary, then it is a/an
(a) acute triangle
(b) equilateral triangle
(c) isosceles triangle
(d) right angled triangle
Answer: (d) right angled triangle
In simple words: Complementary angles sum to 90 degrees. Subtracting that from the triangle's 180-degree total leaves 90 degrees for the third angle, making it a right-angled triangle.

Exam Tip: "Complementary" means summing to \( 90^\circ \), whereas "supplementary" means summing to \( 180^\circ \); never confuse the two terms.

Question 20. (iv) From the fig. \( \angle A + \angle B + \angle C + \angle D + \angle E + \angle F = \)
(a) 90°
(b) 180°
(c) 270°
(d) 360°
Answer: (d) 360°
In simple words: The six-pointed star consists of two overlapping triangles: triangle ACE and triangle BDF. Each triangle contains 180 degrees, so together they sum to 360 degrees.

Exam Tip: Decompose star-like polygons into their fundamental component triangles to quickly determine the total sum of their vertices.