Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 2 Set 2.2 Parallel Lines Solutions

Question 1. In the given figure, y = 108° and x = 71°. Are the lines m and n parallel? Justify?

ℹ️ चित्र व्याख्या (Diagram Explanation): एक तिर्यक रेखा 'l' दो अन्य रेखाओं 'm' और 'n' को काट रही है। रेखा 'm' के ऊपर तिर्यक रेखा के साथ बना कोण 'x' है और रेखा 'n' के नीचे तिर्यक रेखा के साथ बना कोण 'y' है।
Answer:
Solution:
y = 108°, x = 71° ...[Given]
x + y = 71° + 108°
= 179°
∴ x + y = 180°
.. The angles x and y are not supplementary.
.. The angles do not satisfy the interior angles test for parallel lines
.. line m and line n are not parallel lines.
In simple words: If two lines are parallel, the sum of their interior angles on the same side of the transversal should be 180 degrees. Since x + y is 179 degrees and not 180 degrees, the lines m and n are not parallel.

🎯 Exam Tip: Remember the conditions for parallel lines: interior angles are supplementary, alternate interior angles are equal, and corresponding angles are equal. Justification is key to scoring marks.

Question 2. In the given figure, if ∠a = ∠b then prove that line l || line m.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक तिर्यक रेखा 'n' दो रेखाओं 'l' और 'm' को काट रही है। रेखा 'l' पर तिर्यक रेखा के एक ही तरफ के ऊपरी कोण को 'a' और रेखा 'm' पर तिर्यक रेखा के उसी तरफ के निचले कोण को 'b' दर्शाया गया है।
Answer:
Given: ∠a ≈ ∠b
To prove: line l || line m
Solution:
Proof:
ℹ️ चित्र व्याख्या (Diagram Explanation): वही आरेख जिसमें रेखा 'l' पर तिर्यक रेखा के साथ बने कोण 'a' के ठीक नीचे ऊर्ध्वाधर सम्मुख कोण को 'c' के रूप में दर्शाया गया है। कोण 'b' अपनी मूल स्थिति में है।
consider ∠c as shown in the figure ∠a ≈ ∠c ........ (i) [Vertically opposite angles]
But, ∠a ≈ ∠b (ii) [Given]
∴ ∠b = ∠c [From (i) and (ii)]
But, ∠b and ∠c are corresponding angles on lines l and m when line n is the transversal.
.. line l || line m. [Corresponding angles test]
In simple words: By using the vertically opposite angles property, we establish that angle a is equal to angle c. Since it is given that angle a is equal to angle b, it implies that angle b is also equal to angle c. As angles b and c are corresponding angles and are equal, lines l and m are parallel.

🎯 Exam Tip: Clearly state the given, to prove, and provide a step-by-step proof with reasons. Correctly identifying angle relationships is crucial.

Question 3. In the given figure, if ∠a ≈ ∠b and ∠x = ∠y, then prove that line l || line n.
ℹ️ चित्र व्याख्या (Diagram Explanation): तीन रेखाएँ 'l', 'm' और 'n' हैं, जिन्हें एक तिर्यक रेखा 'k' और दूसरी तिर्यक रेखा 'PQ' काट रही है। रेखा 'l' और 'm' के बीच तिर्यक रेखा 'k' द्वारा बनाए गए संगत कोण 'a' और 'b' हैं। रेखा 'm' और 'n' के बीच तिर्यक रेखा 'PQ' द्वारा बनाए गए एकांतर आंतरिक कोण 'x' और 'y' हैं।
Answer:
Given: ∠a ≈ ∠b and ∠x = ∠y
To prove: line l || line n
Solution:
Proof:
ℹ️ चित्र व्याख्या (Diagram Explanation): एक तिर्यक रेखा 'k' द्वारा काटी गई दो रेखाएँ 'l' और 'm' दिखाई गई हैं, जहाँ संगत कोण 'a' और 'b' चिह्नित हैं।
∠a = ∠b [Given]
But, ∠a and ∠b are corresponding angles on lines l and m when line k is the transversal.
∴ line l || line m ....(i) [Corresponding angles test]
∠x = ∠y [Given]
But, ∠x and ∠y are alternate angles on lines m and n when seg PQ is the transversal,
ℹ️ चित्र व्याख्या (Diagram Explanation): तीन रेखाएँ 'l', 'm', 'n' दिखाई गई हैं। तिर्यक रेखा 'k' रेखा 'l' और 'm' को काटती है, जिसमें कोण 'a' और 'b' चिह्नित हैं। तिर्यक रेखा 'PQ' रेखा 'm' और 'n' को काटती है, जिसमें एकांतर आंतरिक कोण 'x' और 'y' चिह्नित हैं।
∴ line m || line n ......(ii) [Alternate angles test]
∴ From (i) and (ii),
line l || line m || line n
i.e., line l || line n
In simple words: Given that ∠a = ∠b (corresponding angles), lines l and m are parallel. Also, given that ∠x = ∠y (alternate interior angles), lines m and n are parallel. Since line l is parallel to line m, and line m is parallel to line n, it logically follows that line l is parallel to line n.

🎯 Exam Tip: Break down the problem into smaller parts. First, prove two lines are parallel, then use that result to prove the next pair. Clearly state the theorems used for each step.

Question 4. In the given figure, if ray BA || ray DE, ∠C = 50° and ∠D = 100°. Find the measure of ∠ABC.
(Hint: Draw a line passing through point C and parallel to line AB.)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक आकृति जिसमें एक कोण ABC है। एक किरण BA और एक किरण DE समानांतर हैं। कोण BCD 50° है और कोण CDE 100° है।
Answer:
Solution:
Draw a line FG passing through point C and parallel to line AB
line FG || ray BA .......(i) [Construction]
Ray BA || ray DE ....(ii) [Given]
line FG || ray BA || ray DE ...(iii) [From (i) and (ii)]
line FG||rayDE [From (iii)]
and seg DC is their transversal
∴ ∠ DCF = ∠ EDC [Alternate angles]
.: ∠ DCF = 100° [: ∠D = 100°]
Now, ∠ DCF = ∠ BCF + ∠ BCD [Angle addition property]
.. 100° = ∠BCF + 50°
.. 100° – 50° = ∠BCF
∴ ∠BCF = 50° ....(iv)
Now, line FG || ray BA and seg BC is their transversal.
ℹ️ चित्र व्याख्या (Diagram Explanation): मूल आकृति में, बिंदु C से गुजरती हुई एक अतिरिक्त रेखा FG खींची गई है जो किरण BA के समानांतर है।
∴ ∠ABC + ∠BCF = 180° [Interior angles]
∴ ∠ABC + 50° = 180° [From (iv)]
.. ∠ABC = 180°-50°
.. ∠ABC = 130°
In simple words: By constructing a line parallel to BA through C, we can use alternate angles to find ∠DCF. Then, using the angle addition property, we find ∠BCF. Finally, applying the property of interior angles being supplementary, we can calculate ∠ABC.

🎯 Exam Tip: Auxiliary constructions are often helpful in geometry problems. Clearly state your construction and use it to establish angle relationships. Remember to use interior angles theorem for parallel lines.

Question 5. In the given figure, ray AE || ray BD, ray AF is the bisector of ∠EAB and ray BC is the bisector of ∠ABD. Prove that line AF || line BC.
ℹ️ चित्र व्याख्या (Diagram Explanation): दो समानांतर किरणें AE और BD हैं। एक तिर्यक रेखा AB इन्हें काटती है। किरण AF, कोण EAB को समद्विभाजित करती है, और किरण BC, कोण ABD को समद्विभाजित करती है। कोण EAF को 'x' और कोण FAB को भी 'x' के रूप में दर्शाया गया है। कोण ABC को 'y' और कोण CBD को भी 'y' के रूप में दर्शाया गया है।
Answer:
Given: Ray AE || ray BD, and ray AF and ray BC are the bisectors of ∠EAB and ∠ABD respectively.
To prove: line AF || line BC
Solution:
Proof:
Ray AE || ray BD and seg AB is their transversal.
∴ ∠EAB = ∠ABD ....(i) [Alternate angles]
∠FAB = \( \frac{1}{2} \) ∠EAB [Ray AF bisects ∠EAB]
∴ 2∠FAB = ∠EAB .....(ii)
∠CBA = \( \frac{1}{2} \) ∠ABD [Ray BC bisects ∠ABD]
∴ 2∠CBA = ∠ABD ...(iii)
∴ 2∠FAB = 2∠CBA [From (i), (ii) and (iii)]
∴ ∠FAB = ∠CBA
But, ∠FAB and ∠ABC are alternate angles on lines AF and BC when seg AB is the transversal.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक तिर्यक रेखा AB द्वारा काटी गई दो रेखाएँ AF और BC हैं। कोण FAB को 'x' और कोण CBA को 'y' के रूप में चिह्नित किया गया है, यह दर्शाते हुए कि वे एकांतर आंतरिक कोण हैं।
∴ line AF || line BC [Alternate angles test]
In simple words: Since AE || BD, the alternate interior angles ∠EAB and ∠ABD are equal. As AF and BC are angle bisectors, ∠FAB is half of ∠EAB, and ∠CBA is half of ∠ABD. Because the full angles are equal, their halves must also be equal (∠FAB = ∠CBA). Since these are alternate interior angles for lines AF and BC with transversal AB, it proves that AF || BC.

🎯 Exam Tip: When dealing with angle bisectors, remember to express the bisected angles as half of the original angle. Alternate interior angles are a powerful tool for proving lines parallel.

Question 6. A transversal EF of line AB and line CD intersects the lines at points P and Q respectively. Ray PR and ray QS are parallel and bisectors of ∠BPQ and ∠PQC respectively. Prove that line AB || line CD.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक तिर्यक रेखा EF दो रेखाओं AB और CD को क्रमशः P और Q पर काटती है। किरण PR कोण BPQ को समद्विभाजित करती है और किरण QS कोण PQC को समद्विभाजित करती है। PR और QS भी एक-दूसरे के समानांतर हैं।
Answer:
Solution:
Proof:
ℹ️ चित्र व्याख्या (Diagram Explanation): वही आरेख जिसमें रेखा EF, AB और CD को P और Q पर काटती है। PR, कोण BPQ का समद्विभाजक है और QS, कोण PQC का समद्विभाजक है। PR और QS समानांतर दर्शाए गए हैं।
Ray PR || ray QS and seg PQ is their transversal.
∴ ∠RPQ = ∠SQP ....(i) [Alternate angles]
∠RPQ = \( \frac{1}{2} \) ∠BPQ .... (ii) [Ray PR bisects ∠BPQ]
∠SQP = \( \frac{1}{2} \) ∠PQC [Ray QS bisects ∠PQC]
∴ \( \frac{1}{2} \) ∠BPQ = \( \frac{1}{2} \) ∠PQC
∴ ∠BPQ = ∠PQC
But, ∠BPQ and ∠PQC are alternate angles on lines AB and CD when line EF is the transversal.
∴ line AB || line CD [Alternate angles test]
In simple words: Given that PR || QS, their alternate interior angles (∠RPQ and ∠SQP) are equal. Since PR and QS are angle bisectors, we can write ∠BPQ = 2∠RPQ and ∠PQC = 2∠SQP. Substituting the equality of the bisected angles leads to ∠BPQ = ∠PQC. As these are alternate interior angles for lines AB and CD, the lines AB and CD must be parallel.

🎯 Exam Tip: Understand the properties of angle bisectors and alternate interior angles thoroughly. Writing clear, logical steps with proper reasons is essential for a complete proof.

Maharashtra Board Class 9 Maths Chapter 2 Parallel Lines Practice Set 2.2 Intext Questions And Activities

Question 1. In the given figure, how will you decide whether line l and line m are parallel or not? (Textbook pg. no. 19)
ℹ️ चित्र व्याख्या (Diagram Explanation): दो रेखाएँ, 'l' और 'm', एक ही तल में दिखाई गई हैं और वे एक-दूसरे को कहीं भी काटती हुई नहीं दिख रही हैं।
Answer:
In the figure, we observe that line l and line m are coplanar and do not intersect each other.
∴ Line l and line m are parallel lines.
In simple words: In geometry, two distinct lines in the same plane that never intersect are defined as parallel lines. Based on the visual representation, lines l and m fit this definition.

🎯 Exam Tip: Remember the basic definition of parallel lines: they are coplanar and never intersect. This fundamental concept is crucial for understanding more complex theorems.