2. Parallel Lines

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

Practice Set 2.2 : Problem 1 : Figure 2.18 : Textbook Page 21

Solution:

\(\angle x + \angle y\) = 71° + 108° ... (Given)

∴ \(\angle x + \angle y\) = 179°

∴ \(\angle x + \angle y\) ≠ 180°

But, x and y are interior angles and they are not supplementary.

∴ line m and line n are not parallel.

2. In figure 2.19, if \(\angle a\) \(\cong\) \(\angle b\) then prove that line l | | line m.

Practice Set 2.2 : Problem 2 : Textbook page 21 : Figure 2.19

Proof:

Take \(\angle c\) as shown.

Now, \(\angle a\) \(\cong\) \(\angle b\) ... (Given) ... (i)
And, \(\angle a\) \(\cong\) \(\angle c\) ... (Vertically Opposite Angles) ... (ii)

From (i) and (ii),
\(\angle b\) \(\cong\) \(\angle c\)
∴ line l | | line m ... (Corresponding Angles test)

3. In figure 2.20, if \(\angle a \cong \angle b\) and \(\angle x \cong \angle y\) then prove that line l | | line m.

Practice Set 2.2 : Problem 3 : Figure 2.20 : Textbook Page 21

Proof:

\(\angle a \cong \angle b\) ... (Given)
∴ line l | | line m
... (Corresponding Angles Test) ... (i)

Also, \(\angle x \cong \angle y\) ... (Given)
∴ line m | | line n
... (Alternate Angles Test) ... (ii)

From (i) and (ii):
∴ line l | | line n ... (Transitivity)

4. In figure 2.21, if ray BA | | ray DE, \(\angle\)C = 50° and \(\angle\)D = 100°. Find the measure of \(\angle\)ABC.
(Hint : Draw a line passing through point C and parallel to line AB.)

Solution:

Practice Set 2.2 : Problem 4 : Textbook Page 21 : Figure 2.21

Draw line PQ | | line AB passing through point C such that P – C – Q.

Now, line AB | | line PQ ... (Construction)
and line AB | | line DE ... (Given)
∴ line PQ | | line DE ... (Transitivity)
Consider transversal CD.
\(\angle\)DCQ + \(\angle\)EDC = 180°
... (Interior Angles Theorem)
∴\(\angle\)DCQ + 100° = 180° ... (Given)
∴\(\angle\)DCQ = 180° − 100°
∴\(\angle\)DCQ = 80° ... (i)

Also, \(\angle\)BCQ = \(\angle\)DCQ + \(\angle\)DCB ... (Angle Addition Property)
∴ \(\angle\)BCQ = 80° + 50°
∴ \(\angle\)BCQ = 130° ... (ii)

Finally, line AB | | line PQ ... (Construction)
∴ \(\angle\)ABC = \(\angle\)BCQ ... (Alternate Angles)
But, \(\angle\)BCQ = 130° ... [From (ii)]
∴ \(\angle\)ABC = 130°

5. In figure 2.22, ray AE | | ray BD, ray AF is the bisector of \(\angle\)EAB and ray BC is the bisector of \(\angle\)ABD. Prove that line AF | | line BC.

Proof:

Practice Set 2.2 : Problem 5 : Figure 2.22 : Textbook page 22

Ray AF is the bisector of \(\angle\)EAB. ... (Given)
∴ \(\angle\)EAF \(\cong\) \(\angle\)BAF = x ... (i)

Also, ray BC is the bisector of \(\angle\)ABD. ... (Given)
∴ \(\angle\)DBC \(\cong\) \(\angle\)ABC = y ... (ii)

Now, ray AE | | ray BD ... (Given)
Consider transversal AB.
\(\angle\)EAB \(\cong\) \(\angle\)ABD ... (Alternate Angles) ... (iii)
∴ \(\angle\)EAF + \(\angle\)BAF = \(\angle\)DBC + \(\angle\)ABC
... (Angle Addition Property)
∴ x + x = y + y ... [From (i) and (ii)]
∴ 2x = 2y
∴ x = y
∴ \(\angle\)BAF = \(\angle\)ABC ... [From (i) and (ii)]
∴ line AF | | line BC ... (Alternate Angles Test)

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 \(\angle\)BPQ and \(\angle\)PQC respectively. Prove that line AB | | line CD.

Proof:

Practice Set 2.2 : Problem 6 : Figure 2.21 : Textbook Page 22

ray PR | | ray QS ... (Given)
Consider transversal EF.
\(\angle\)QPR \(\cong\) \(\angle\)SQP ... (Alternate Angles)
Multiplying both sides by 2,
2\(\angle\)QPR = 2\(\angle\)SQP ... (i)
But, ray PR and ray QS are the bisectors of \(\angle\)BPQ and \(\angle\)PQC respectively. ... (Given)
∴ 2\(\angle\)QPR = \(\angle\)BPQ and 2\(\angle\)SQP = \(\angle\)PQC ... (ii)

From (i) and (ii),
\(\angle\)BPQ \(\cong\) \(\angle\)PQC
∴ line AB | | line CD ... (Alternate Angles Test)

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

Solution:

2. In figure 2.19, if \(\angle a\) \(\cong\) \(\angle b\) then prove that line l | | line m.

Proof:

3. In figure 2.20, if \(\angle a \cong \angle b\) and \(\angle x \cong \angle y\) then prove that line l | | line m.

Proof:

4. In figure 2.21, if ray BA | | ray DE, \(\angle\)C = 50° and \(\angle\)D = 100°. Find the measure of \(\angle\)ABC. (Hint : Draw a line passing through point C and parallel to line AB.)

Solution:

5. In figure 2.22, ray AE | | ray BD, ray AF is the bisector of \(\angle\)EAB and ray BC is the bisector of \(\angle\)ABD. Prove that line AF | | line BC.

Proof:

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 \(\angle\)BPQ and \(\angle\)PQC respectively. Prove that line AB | | line CD.

Proof:

4. In figure 2.21, if ray BA | | ray DE, \(\angle\)C = 50° and \(\angle\)D = 100°. Find the measure of \(\angle\)ABC.
(Hint : Draw a line passing through point C and parallel to line AB.)