Std 9 Mathematics Part 2 · Chapter 2

2. Parallel Lines

Solutions to Practice Sets 2.1, 2.2 and Problem Set 2, with figures and proofs. Maharashtra State Board (Balbharati), Std 9 Mathematics Part 2 (Geometry).

Practice Set 2.1

1In figure 2.5, line RP ∥ line MS and line DK is their transversal. ∠DHP = 85°. Find the measures of (i) ∠RHD (ii) ∠PHG (iii) ∠HGS (iv) ∠MGK.

Fig. 2.5 — line RP ∥ line MS, transversal DK

∠RHD + ∠DHP = 180° (angles in a linear pair)
∠RHD + 85° = 180°
∴ ∠RHD = 95°
∠PHG + ∠DHP = 180° (angles in a linear pair)
∠PHG = 180° − 85° = 95°
∠HGS = ∠DHP (corresponding angles, line RP ∥ line MS)
∴ ∠HGS = 85°
∠MGK = ∠HGS (vertically opposite angles)
∴ ∠MGK = 85°

2In 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.

Fig. 2.6 — line p ∥ line q; line l and line m are transversals

∠a : ∠a and the 110° angle form a linear pair.
∠a + 110° = 180° ∴ ∠a = 70°

∠b : At the intersection of line p and line l, the angle vertically opposite to ∠a is also 70°. This angle and ∠b are corresponding angles for line p ∥ line q with transversal l.
∴ ∠b = 70°

∠c : ∠c and the 115° angle are corresponding angles for line p ∥ line q with transversal m.
∴ ∠c = 115°

∠d : ∠d and the 115° angle form a linear pair.
∠d = 180° − 115° = 65°

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

Fig. 2.7 — line l ∥ line m, line n ∥ line p

Let ∠k be the angle at the intersection of line p and line m that lies above line m and to the right of line p.

∠k = 45° (corresponding angles, line l ∥ line m, transversal p)

∠a : ∠a + ∠k = 180° (angles in a linear pair)
∠a = 180° − 45° = 135°

∠b : ∠b = ∠a (vertically opposite angles)
∴ ∠b = 135°

∠c : At the intersection of line n and line m, the angle above line m and to the right of line n corresponds to ∠k (line n ∥ line p, transversal m), so it is 45°. ∠c is vertically opposite to it.
∴ ∠c = 45°

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

Fig. 2.8 — ray QP ∥ ray YX, ray QR ∥ ray YZ; line XY produced meets ray QR at T

Given: Ray QP ∥ ray YX and ray QR ∥ ray YZ.

To prove: ∠PQR ≅ ∠XYZ

Construction: Produce seg XY to meet ray QR at point T.

Line QP ∥ line XT and line QR is a transversal.
∴ ∠PQR ≅ ∠XTR ...... (I) corresponding angles

Line YZ ∥ line TR and line XT is a transversal.
∴ ∠XYZ ≅ ∠XTR ...... (II) corresponding angles

∴ ∠PQR ≅ ∠XYZ ...... from (I) and (II)

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

Fig. 2.9 — line AB ∥ line CD, transversal PQ, ∠BRT = 105°

∠ART + ∠BRT = 180° (angles in a linear pair)
∠ART = 180° − 105° = 75°
∠CTQ = ∠ART (corresponding angles, line AB ∥ line CD)
∴ ∠CTQ = 75°
∠DTQ = ∠BRT (corresponding angles)
∴ ∠DTQ = 105°
∠PRB + ∠BRT = 180° (angles in a linear pair)
∠PRB = 180° − 105° = 75°

Practice Set 2.2

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

Fig. 2.18

∠x and ∠y are interior angles on the same side of transversal l.

∠x + ∠y = 71° + 108° = 179°

The sum is not 180°, so the interior angles are not supplementary. By the interior angles test, line m and line n are not parallel.

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

Fig. 2.19 — ∠c is marked for the proof

Given: ∠a ≅ ∠b

To prove: line l ∥ line m

Let ∠c be the angle vertically opposite to ∠b, as shown.

∠b ≅ ∠c ...... (I) vertically opposite angles

∠a ≅ ∠b ...... (II) given

∴ ∠a ≅ ∠c ...... from (I) and (II)

But ∠a and ∠c are corresponding angles formed by transversal n of line l and line m.

∴ line l ∥ line m ...... corresponding angles test

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

Fig. 2.20

Given: ∠a ≅ ∠b and ∠x ≅ ∠y

To prove: line l ∥ line n

∠a and ∠b are corresponding angles formed by transversal K of line l and line m, and ∠a ≅ ∠b ...... given

∴ line l ∥ line m ...... (I) corresponding angles test

∠x and ∠y are alternate angles formed by a transversal of line m and line n, and ∠x ≅ ∠y ...... given

∴ line m ∥ line n ...... (II) alternate angles test

From (I) and (II), line l and line n are both parallel to line m.

∴ line l ∥ line n ...... lines parallel to the same line are parallel to each other

The last step is Corollary II of this chapter. Its proof: if line l and line n were not parallel, they would meet at a point, giving two different lines through that point both parallel to line m. That contradicts the fact that through a point outside a line, only one line parallel to it can be drawn.

4In 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.)

Fig. 2.21

Construction: line FG through C, parallel to ray BA

Construction: Draw line FG through point C parallel to line AB, with F on the side of A and G on the side of E.

Ray DE ∥ ray BA and line FG ∥ ray BA, so line FG ∥ ray DE.

Line FG ∥ ray DE and seg CD is a transversal.
∴ ∠GCD + ∠CDE = 180° ...... interior angles
∠GCD + 100° = 180° ∴ ∠GCD = 80°

∠BCG = ∠BCD + ∠DCG = 50° + 80° = 130°

Line FG ∥ ray BA and seg BC is a transversal.
∴ ∠ABC = ∠BCG ...... alternate angles

∴ ∠ABC = 130°

5In 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.

Fig. 2.22 — ∠EAF = ∠FAB = x, ∠ABC = ∠CBD = y

Given: Ray AE ∥ ray BD; ray AF bisects ∠EAB; ray BC bisects ∠ABD.

To prove: line AF ∥ line BC

∠EAB = 2x and ∠ABD = 2y ...... given bisectors

Ray AE ∥ ray BD and seg AB is a transversal.
∴ ∠EAB = ∠ABD ...... alternate angles

∴ 2x = 2y, that is, x = y

∴ ∠FAB = ∠ABC

But ∠FAB and ∠ABC are alternate angles formed by transversal AB of line AF and line BC.

∴ line AF ∥ line BC ...... alternate angles test

6A 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.

Fig. 2.23

Given: Ray PR ∥ ray QS; ray PR bisects ∠BPQ; ray QS bisects ∠PQC.

To prove: line AB ∥ line CD

Ray PR ∥ ray QS and seg PQ is a transversal.
∴ ∠RPQ = ∠SQP ...... (I) alternate angles

∠BPQ = 2∠RPQ ...... ray PR bisects ∠BPQ

∠PQC = 2∠SQP ...... ray QS bisects ∠PQC

∴ ∠BPQ = ∠PQC ...... from (I)

But ∠BPQ and ∠PQC are alternate angles formed by transversal EF of line AB and line CD.

∴ line AB ∥ line CD ...... alternate angles test

Problem Set 2

1Select 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 — (C) 180°
Interior angles on the same side are supplementary.
(ii) The number of angles formed by a transversal of two lines is — (C) 8
Four angles are formed at each of the two points of intersection.
(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°
Corresponding angles are congruent.
(iv) In ΔABC, ∠A = 76°, ∠B = 48°, ∴ ∠C = — (B) 56°
∠C = 180° − (76° + 48°) = 180° − 124° = 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 — (C) 75°
Alternate angles are congruent.

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. Draw a figure showing all these rays and write — (i) a pair of complementary angles (ii) a pair of supplementary angles (iii) a pair of congruent angles.

Ray PQ ⊥ ray PR; B lies inside ∠QPR, A lies outside it; ray PB ⊥ ray PA

Complementary angles: ∠QPB and ∠BPR, since ∠QPB + ∠BPR = ∠QPR = 90°. (∠APQ and ∠QPB are also complementary, since together they form ∠APB = 90°.)
Supplementary angles: ∠QPR and ∠APB, since 90° + 90° = 180°.
Congruent angles: ∠APQ ≅ ∠BPR. Both are complements of ∠QPB: ∠APQ = 90° − ∠QPB and ∠BPR = 90° − ∠QPB. (∠QPR ≅ ∠APB is another pair, as both are right angles.)

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

Line l ∥ line m; line n ⊥ line l at M and meets line m at N

Given: Line l ∥ line m; line n ⊥ line l.

To prove: line n ⊥ line m

Line n ⊥ line l ∴ ∠a = 90° ...... given

Line l ∥ line m and line n is a transversal.
∴ ∠b = ∠a ...... corresponding angles

∴ ∠b = 90°

∴ line n ⊥ line m

4In 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.

Fig. 2.24

∠x = 130° ...... vertically opposite to the 130° angle
∠y = 50° ...... vertically opposite to the 50° angle

∠x and the 50° angle are interior angles on the same side of the transversal.
∠x + 50° = 130° + 50° = 180°

∴ the interior angles are supplementary.
∴ line l ∥ line m ...... interior angles test

∴ ∠x = 130°, ∠y = 50°

The same conclusion follows from ∠y and the 130° angle: they are exterior angles on the same side of the transversal and add up to 180°.

5Line 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.)

Fig. 2.25 — line AB ∥ line CD ∥ line EF, transversal QP

Let the angle at line EF that lies above line EF and to the left of the transversal be ∠w.
∠w + ∠z = 180° ...... angles in a linear pair

∠y = ∠w ...... corresponding angles, line CD ∥ line EF
∴ ∠y + ∠z = 180° ...... (I)

y : z = 3 : 7, so let ∠y = 3k and ∠z = 7k.
3k + 7k = 180° ∴ 10k = 180° ∴ k = 18°
∴ ∠y = 54° and ∠z = 126°

∠x and ∠y are interior angles on the same side of the transversal of line AB ∥ line CD.
∴ ∠x + ∠y = 180°
∠x = 180° − 54° = 126°

∴ ∠x = 126°

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

Fig. 2.26 — line q ∥ line r, transversal p

f : ∠a + ∠b = 180° (angles in a linear pair) ∴ ∠b = 180° − 80° = 100°
∠f = ∠b (corresponding angles, line q ∥ line r)
∴ f = 100°

g : ∠c = ∠a = 80° (vertically opposite angles)
∠g = ∠c (corresponding angles)
∴ g = 80°

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

Fig. 2.27

Given: Line AB ∥ line CF and line BC ∥ line ED.

To prove: ∠ABC = ∠FDE

Line AB ∥ line CF and seg BC is a transversal.
∴ ∠ABC = ∠BCD ...... (I) alternate angles

Line BC ∥ line ED and line CF is a transversal.
∴ ∠BCD = ∠EDF ...... (II) corresponding angles

∴ ∠ABC = ∠FDE ...... from (I) and (II)

8In 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.

Fig. 2.28

Given: Line AB ∥ line CD, transversal PS meets them at Q and R. Ray QX, ray QY, ray RX and ray RY bisect ∠AQR, ∠BQR, ∠QRC and ∠QRD respectively.

To prove: □QXRY is a rectangle.

∠AQR + ∠BQR = 180° ...... angles in a linear pair
∴ ½∠AQR + ½∠BQR = 90°
∴ ∠XQR + ∠RQY = 90° ...... rays QX and QY are bisectors
∴ ∠XQY = 90° ...... (I)

Similarly, ∠QRC + ∠QRD = 180°
∴ ∠QRX + ∠QRY = 90°
∴ ∠XRY = 90° ...... (II)

∠BQR + ∠QRD = 180° ...... interior angles, line AB ∥ line CD
∴ ½∠BQR + ½∠QRD = 90°
∴ ∠YQR + ∠QRY = 90°

In ΔQYR, ∠QYR = 180° − (∠YQR + ∠QRY) = 180° − 90° = 90° ...... (III)

∠AQR + ∠QRC = 180° ...... interior angles
∴ ∠XQR + ∠QRX = 90°
In ΔQXR, ∠QXR = 180° − 90° = 90° ...... (IV)

From (I), (II), (III) and (IV), all four angles of □QXRY are right angles.

∴ □QXRY is a rectangle.

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