## Chapter 2 – Parallel Lines

**Practice set 2.1**

**1. In figure 2.5, line RP || line MS and line DK is their transversal. ∠DHP = 85°. Find the measures of following angles.**

**(i) ∠RHD **

**(ii) ∠PHG**

**(iii) ∠HGS **

**(iv) ∠MGK**

**Given:**

line RP || line MS and line DK is their transversal

∠DHP = 85°

**To find:**

(i) ∠RHD

(ii) ∠PHG

(iii) ∠HGS

(iv) ∠MGK

**Solution:**

Line RP || line MS and line DK is their transversal,

(i) ∠DHP + ∠RHD = 180° …*[Angles in linear pair]*

∴ 85 + ∠RHD = 180

∴ ∠RHD = 180 – 85

∴ ∠RHD = 95°

(ii) ∠PHG ≅ ∠RHD …*[Vertically opposite angles]*

∵ ∠RHD = 95°

∴ ∠PHG = 95°

(iii) ∠DHP ≅ ∠HGS …*[Corresponding angles theorem]*

∵ ∠DHP = 85°

∴ ∠HGS = 85°

(iv) ∠HGS ≅ ∠MGK …*[Vertically opposite angles]*

∵ ∠HGS = 85°

∴ ∠MGK = 85°

**Ans:** The measures are;

(i) ∠RHD = 95⁰

(ii) ∠PHG = 95⁰

(iii) ∠HGS = 85⁰

(iv) ∠MGK = 85⁰

**2. In figure 2.6, line p || line q and line l and line m are transversals. Measures of some angles are shown. Hence find the measures of ∠a, ∠b, ∠c, ∠d.**

**Given:**

Line p || line q and line l and line m are transversals

Measures of two angles are 100⁰ and 115⁰.

**To find:**

∠a

∠b

∠c

∠d

**Solution:**

∠a + 110 = 180 …*[Angles in linear pair]*

∴ ∠a = 180 – 110

∴ ∠a = 70°

Line p || line q and line l is their transversal

∠b ≅ ∠a …*[Exterior Alternate angles theorem]*

∵ ∠a = 70⁰

∴ ∠b = 70⁰

Line p || line q and line l is their transversal

∠c = 115° …*[Corresponding angles theorem]*

∠d + 115 = 180 …*[Angles in linear pair]*

∴ ∠d = 180 – 115

∴ ∠d = 65°

**Ans:** The measures are;

∠a = 70⁰

∠b = 70⁰

∠c = 115⁰

∠d = 65⁰

**3. In figure 2.7, line l || line m and line n || line p. Find ∠a, ∠b, ∠c from the given measure of an angle.**

**Given:**

line l || line m

line n || line p

One angle is 45⁰

**To find:**

∠a

∠b

∠c

**Solution:**

Let us mark an angle ‘d’ as shown in the figure,

∴ ∠d = 45° …*[Vertically opposite angles]*

Line p || line q and line l is their transversal,

∠d + ∠a = 180° …*[Interior angles theorem]*

∴ 45 + ∠a = 180

∴ ∠a = 180 – 45

∴ ∠a = 135°

∴ ∠b = ∠a …*[Vertically opposite angles]*

∵ ∠a = 135°

∴ ∠b = 135°

Line n || line p and line m is their transversal,

∠c ≅ ∠b …*[Corresponding angles theorem]*

∵ ∠b = 135°

∴ ∠c = 135°

**Ans:** The measures are;

∠a = 135⁰

∠b = 135⁰

∠c = 135⁰

**4*. In figure 2.8, sides of ∠PQR and ∠XYZ are parallel to each other. Prove that, ∠PQR ≅ ∠XYZ**

**Given:**

Line PQ || line XY

Line QR || line YZ

**To prove:**

∠PQR ≅ ∠XYZ

**Construction:**

Extend ray XY to intersect ray QR at point M, such that Q – M – R

**Solution:**

Line PQ || line XM and QR is their transversal …*[Given, X – Y – M]*

∠PQR ≅ ∠XMR …(i) *[Corresponding angles theorem]*

Line YZ || line QR and XM is their transversal …*[Given]*

∠XYZ ≅ ∠XMR …(ii) *[Corresponding angles theorem]*

∴ ∠PQR ≅ ∠XYZ …*[From (i) and (ii)]*

**Hence proved.**

**5. In figure 2.9, line AB || line CD and line PQ are transversal. Measure of one of the angles is given. Hence find the measures of the following angles. ****(i) ∠ART ****(ii) ∠CTQ ****(iii) ∠DTQ ****(iv) ∠PRB**

**Given:**

Line AB || line CD and line PQ are transversal.

∠BRT = 105⁰

**To find:**

(i) ∠ART

(ii) ∠CTQ

(iii) ∠DTQ

(iv) ∠PRB

**Solution:**

(i) ∠BRT + ∠ART = 180° …*[Angles in linear pair]*

∴ 105 + ∠ART = 180

∴ ∠ART = 180 – 105

∴ ∠ART = 75°

(ii) Line AB || line CD and line PQ are transversal.

∠ART ≅ ∠CTQ …*[Corresponding angles theorem]*

∵ ∠ART = 75°

∴ ∠CTQ = 75°

(iii) Line AB || line CD and line PQ are transversal

∠BRT ≅ ∠DTQ …*[Corresponding angles theorem]*

∵ ∠BRT = 105⁰

∴ ∠DTQ = 105°

(iv) ∠PRB ≅ ∠ART …*[Vertically opposite angles]*

∵ ∠ART = 75°

∴ ∠PRB = 75°

**Ans:** The measures are;

(i) ∠ART = 75⁰

(ii) ∠CTQ = 75⁰

(iii) ∠DTQ = 105⁰

(iv) ∠PRB = 75⁰

**Practice set 2.2**

**1. In figure 2.18, y = 108° and x = 71° Are the lines m and n parallel? Justify?**

**Given:**

**y = 108°**

**x = 71°**

** **

**To find:**

**Whether the lines m and n are parallel**

** **

**Solution:**

**∠y = 108° …(i)**

**∠x = 71° …(ii)**

** **

**Here, ∠x and ∠y form a pair of interior angles.**

**Hence their sum should be 180⁰**

** **

**Adding (i) and (ii), we get, **

**∠x + ∠y = 71 + 108**

**∴ ∠x + ∠y = 179° ≠ 180⁰**

** **

**Since, ∠x and ∠y are not supplementary,**

**∴ line m is not parallel to line n. **

** **

**Ans:** Line m is not parallel to line n.

**2. In figure 2.19, if ∠a ≅ ∠b then prove that line l || line m.**

**Given:**

**∠a ≅ ∠b**

** **

**To prove:**

**line l || line m**

** **

**Proof:**

**∠a ≅ ∠c …(i) [Vertically opposite angles]**

**∠a ≅ ∠b …(ii) [Given]**

** **

**∴ ∠b ≅ ∠c … [From (i) and (ii)]**

**∴ line l line m … [Corresponding angles test]**

** **

**Hence proved.**

**3. In figure 2.20, if ∠a ≅ ∠b and ∠x ≅ ∠y then prove that line l || line n.**

**Given:**

**∠a ≅ ∠b**

**∠x ≅ ∠y**

** **

**To prove:**

**line l || line n**

** **

**Proof:**

**∠a ≅ ∠b … [Given]**

**∴ line l || line m … (i) [Corresponding angles test]**

** **

**∠x ≅ ∠y … [Given]**

**∴ line n || line m …(ii) [Alternate angles test]**

** **

**∴ line l || line n … [From (i) and (ii)]**

** **

**Hence proved.**

**4. In figure 2.21, 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.)**

**Given:**

**ray BA || ray DE**

**∠C = 50°**

**∠D = 100°**

** **

**To find:**

**∠ABC**

** **

**Construction: **

**Draw a line passing through point C and parallel to line AB.**

**Solution:**

**line AB || line CF. …(i) [Construction]**

**line AB || line DE. …(ii) [Given]**

** **

**∴ line DE || line CF … [From (i) and (ii)]**

** **

**line DE || line CF and DC is their transversal, **

**∠EDC + ∠DCF = 180° … [Interior angles theorem]**

**∴ 100 + ∠DCF = 180**

**∴ ∠DCF = 180 – 100**

**∴ ∠DCF = 80°**

** **

**∠BCF = ∠BCD + ∠DCF … [Angle addition property]**

**∴ ∠BCF = 50 + 80**

**∴ ∠BCF = 130°**

** **

**line AB || line CF and BC is their transversal, **

**∠ABC ≅ ∠BCF … [Alternate angles theorem]**

**∵ ∠BCF = 130°**

**∴ ∠ABC = 130°**

** **

**Ans:** The measure of ∠ABC is 130°.

**5. In figure 2.22, 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.**

**Given:**

**ray AE || ray BD**

**ray AF is the bisector of ∠EAB**

**ray BC is the bisector of ∠ABD**

** **

**To prove:**

**line AF || line BC**

** **

**Proof:**

**∠EAF ≅ ∠BAF = x …(i) [∵ ray AF bisects ∠EAB]**

** **

**∠DBC = ∠ABC = y …(ii) [∵ ray BC bisects ∠ABD]**

** **

**ray AE || ray BD and AB is the transversal … [Given]**

**∠EAB ≅ ∠ABD … [Alternate angles theorem]**

** **

**∴ ∠EAF + ∠BAF = ∠ABC + ∠DBC … [Angles addition property]**

** x + x = y + y … [From (i) and (ii)]**

**∴ 2x = 2y**

**∴ x = y**

** **

**∴ ∠FAB ≅ ∠ABC**

**∴ line AF line BC … [Alternate angles test]**

** **

**Hence proved. **

**6. A transversal EF of line AB and line CD intersects the lines at point P and Q respectively. Ray PR and ray QS are parallel and bisectors of ∠BPQ and ∠PQC respectively. Prove that line AB || line CD.**

**Given:**

**Ray PR || ray QS bisectors of **

**Ray PR bisect ∠BPQ **

**Ray PR bisect ∠PQC **

** **

**To prove:**

**line AB || line CD**

** **

**Proof:**

**Let, **

**∠BPR ≅ ∠QPR = x …(i) [∵ ray PR bisects ∠BPQ]**

** **

**∠CQS ≅ ∠PQS = y …(ii) [∵ ray QS bisects ∠PQC]**

** **

**∴ ∠BPQ = ∠BPR + ∠QPR … [Angles addition property]**

**∴ ∠BPQ = x + x … [From (i)]**

**∴ ∠BPQ = 2x …(iii)**

** **

**Similarly, we will get**

**∴ ∠PQC = 2y …(iv)**

** **

**Ray PR || ray QS and PQ is the transversal … [Given]**

**∠QPR ≅ ∠PQS … [Alternate angles theorem]**

**∴ x = y … [From (i) and (ii)]**

** **

**∴ 2(x) = 2(y) … [Multiplying by 2 on both sides]**

**∴ ∠BPQ = ∠PQC … [From (iii) and (iv)]**

** **

**∴ line AB line CD … [Alternate angles test]**

** **

**Hence proved. **

**Problem Set 2**

**Problem Set 2**

**1. Select the correct alternative and fill in the blanks in the following statements. **

**1. Select the correct alternative and fill in the blanks in the following statements.****(i) If a transversal intersects two parallel lines then the sum of interior angles on the same side of the transversal is ______**

**(A) 0° **

**(B) 90° **

**(C) 180° **

**(D) 360°**

**Ans:** Option (C) : 180°

**Solution:**

**line l || line m and n is the transversal,**

**∠a + ∠b = 180° … [Interior angles theorem]**

** **

**(ii) The number of angles formed by a transversal of two lines is ______**

**(A) 2 **

**(B) 4 **

**(C) 8 **

**(D) 16**

** **

**Ans:** Option (C) : 8

**Solution: **

**lines l || line m and line n is the transversal.**

**The number of angles formed by a transversal of two lines as shown in the figure is 8.**

** **

**(iii) A transversal intersects two parallel lines. If the measure of one of the angles is 40°then the measure of its corresponding angle is ______**

**(A) 40° **

**(B) 140° **

**(C) 50° **

**(D) 180°**

** **

**Ans:** Option (A) : 40°

**Solution:**

**line l || line m and n is the transversal, **

**∠a ≅ ∠b … [Corresponding angles theorem]**

**∵ ∠a = 40°**

**∴ ∠b = 40°**

** **

**(iv) In ∆ABC, ∠A = 76°, ∠B = 48°, ∠C = ______**

**(A) 66° **

**(B) 56° **

**(C) 124° **

**(D) 28°**

** **

**Ans:** Option (B) : 56°

**Solution:**

**In ∆ABC, **

**∠A + ∠B + ∠C = 180° … [Sum of measures of all angles of a triangle is 180°]**

**∴ 76 + 48 + ∠C = 180**

**∴ 124 + ∠C = 180**

**∴ ∠C = 180 – 124**

**∴ ∠C = 56° **

** **

**(v) Two parallel lines are intersected by a transversal. If measure of one of the alternate interior angles is 75° then the measure of the other angle is ______**

**(A) 105° **

**(B) 15° **

**(C) 75° **

**(D) 45°**

** **

**Ans: **Option (C) : 75°

**Solution: **

**line l || line m and n is the transversal**

**∠a ≅ ∠b … [Alternate angles theorem]**

**∵ ∠a = 75°**

**∴ ∠b = 75°**

**2*. Ray PQ and ray PR are perpendicular to each other. Points B and A are in the interior and exterior of ∠QPR respectively. Ray PB and ray PA are perpendicular to each other. **

**2*. Ray PQ and ray PR are perpendicular to each other. Points B and A are in the interior and exterior of ∠QPR respectively. Ray PB and ray PA are perpendicular to each other.**

**(i) A pair of complementary angles **

**Ans:**

**(a) ∠QPB and ∠BPR**

**(b) ∠BPR and ∠RPA**

**(ii) A pair of supplementary angles.**

**Ans:** ∠QPR and ∠BPA

**(iii) A pair of congruent angles.**

**Ans: **

**(a) ∠QPR and ∠BPA**

**(b) ∠QPB and ∠RPA**

**3. Prove that, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also.**

**Given:**

**Line AB || line CD and line EF intersects them at points M and N respectively. **

**line EF ⊥ line AB**

** **

**To prove: **

**line EF ⊥ line CD**

** **

**Proof:**

**line AB line CD and EF is the transversal**

**∠EMB ≅ ∠MND …(i) [Corresponding angles theorem]**

**∵ ∠EMB = 90° …(ii) [Given]**

**∴ ∠MND = 90° … [From (i) and (ii)]**

** **

**∴ line EF ⊥ line CD**

** **

**Hence proved.**

**4. In figure 2.24, measures of some angles are shown. Using the measures find the measures of ∠x and ∠y and hence show that line l || line m.**

**Given:**

**∠AME = 130⁰**

**∠MND = 150⁰**

** **

**To find:**

**∠x and ∠y**

** **

**To prove:**

**line l || line m**

** **

**Solution:**

**∠BMN ≅ ∠AME … [Vertically opposite angles]**

**∴ ∠x = 130°**

** **

**∠CNF = ∠MND … [Vertically opposite angles]**

**∴ ∠y = 50°**

** **

**∠BMN + ∠MND = 130 + 50**

**∴ ∠BMN + ∠MND = 180°**

** **

**But, **

**∠BMN and ∠MND are interior angles**

** **

**∴ line l || line m … [By Interior angles test]**

** **

**Hence proved.**

** **

**Ans:** The measures of ∠x is 130⁰ and ∠y = 50⁰

**5. Line AB || line CD || line EF and line QP is their transversal. If y : z = 3 : 7 then find the measure of ∠x. (See figure 2.25.)**

**Given:**

**Line AB || line CD || line EF and line QP is their transversal**

**y : z = 3 : 7**

** **

**To find:**

**∠x**

** **

**Solution:**

**∠y = ∠z = 3 : 7 … [Given]**

** **

**Let the common multiple be a**

**∴ ∠y = 3a and ∠z = 7a**

** **

**line AB || line EF and PQ is the transversal, **

**∴ ∠x ≅ ∠z … [Alternate angles theorem]**

**∵ ∠z = 7a**

**∴ ∠x = 7a**

** **

**line AB || line CD and PQ is the transversal, **

**∠x + ∠y = 180° … [Interior angles theorem]**

**∴ 7a + 3a = 180**

**∴ 10a = 180**

**∴ a = \(\large \frac {180}{10}\)**

** ∴ a = 18**

** **

**Now,**

**∠x = 7a**

**∴ ∠x = 7 × 18**

**∴ ∠x = 126⁰**

** **

**Ans:** The measure of ∠x is 126⁰.

**6. In figure 2.26, if line q || line r, line p is their transversal and if a = 80° find the values of f and g.**

**Given:**

**line q || line r, line p is their transversal**

**∠a = 80°**

** **

**To find:**

**∠f and ∠g**

** **

**Solution:**

**∠a + ∠b = 180° … [Angles in linear pair]**

**∴ 80 + ∠b = 180**

**∴ ∠b = 100°**

** **

**∠c = ∠a … [Vertically opposite angles**

**∵ ∠a = 80°… [Given]**

**∴ ∠c = 80°**

** **

**line q || line r and line p is their transversal**

**∠f = ∠b … [Corresponding angles theorem]**

**∵ ∠b = 100°**

**∴ ∠f = 100°**

** **

**∠g = ∠c … [Corresponding angles theorem]**

**∵ ∠a = 80⁰**

**∴ ∠g = 80°**

** **

**Ans:** The measures of ∠f is 100° and ∠g is 80°.

**7. In figure 2.27, if line AB || line CF and line BC || line ED then prove that ∠ABC = ∠FDE.**

**Given:**

**line AB || line CF**

**line BC || line ED **

** **

**To prove:**

**∠ABC ≅ ∠FDE**

** **

**Proof:**

**line AB || line CF and BC is the transversal, **

**∠ABC = ∠BCD …(i) [Alternate angles theorem]**

** **

**line BC || line ED and BC is the transversal, **

**∠FDE = ∠BCD …(ii) [Corresponding angles theorem]**

** **

**∴ ∠ABC = ∠FDE … [From (i) and (ii)]**

** **

**Hence proved.**

**8. In figure 2.28, line PS is a transversal of parallel line AB and line CD. If Ray QX, ray QY, ray RX, ray RY are angle bisectors, then prove that □ QXRY is a rectangle.**

**Given:**

**Line AB || line CD and PS is the transversal **

**Ray QX, ray QY, ray RX, ray RY are angle bisectors **

** **

**To prove:**

**□ QXRY is a rectangle**

** **

**Solution:**

**Let, ∠AQX = ∠RQX = a …(i)**

**∠BQY = ∠RQY = b …(ii)**

**∠CRX = ∠QRX = c …(iii)**

**∠DRY = ∠QRY = d …(iv) [∵ Rays QX, QY, RX, RY are the bisectors of ∠AQR, ∠BQR, ∠QRC, ∠QRD respectively]**

** **

**∠AQR + ∠BQR = 180° … [Angles in linear pair]**

**∴ 2a + 2b = 180**

**∴ a + b = 90**

**∴ ∠RQX + ∠RQY = 90° … [From (i) and (ii)]**

**∴ ∠XQY = 90° …(v) [Angles addition property]**

** **

**Similarly, we can prove, **

**c + d = 90⁰**

**∴ ∠QRX + ∠QRY = 90⁰**

**∴ ∠XRY = 90° …(vi)**

** **

**line AB || line CD and PS is the transversal, **

**∠AQR + ∠CRQ = 180° … [Interior angles theorem]**

**∴ 2a + 2c = 180**

**∴ ∠a + ∠c = 90 …(vii)**

** **

**In ∆XQR,**

**∠QXR + ∠XQR + ∠XRQ = 180° … [Sum of measures of all angles of a triangle is 180⁰]**

**∴ ∠QXR + ∠a + ∠c = 180 … [From (i) and (iii)]**

**∴ ∠QXR + 90 = 180 … [From (vii)]**

**∴ ∠QXR = 90° …(viii)**

** **

**Similarly, we can prove**

**∠QYR = 90° …(ix)**

** **

**In □ QXRY,**

**∴ ∠XQY ≅ ∠QXR ≅ ∠XRY ≅ ∠RYQ = 90° … [From (v), (vi), (viii) and (ix)]**

**∴ QXRY is a rectangle … [By definition]**

** **

**Hence proved. **