We have ABIIED and BC||EF
(i) BC is transversal
∠DGC = ∠ABC [corresponding angles]
But ∠ABC = 70°
∴ ∠DGC = 70°
(ii) ED is a transversal to BC|lEF
∴ ∠DEF = ∠DGC [corresponding]
∠DGC = 70°
∠DEF = 70°

Step 1: Draw a line. Mark two points A and B on it so that AB = 6 cm.

Step 2: Using compass with A as center and radius more than half of the length of AB, draw two arcs of same length, one above AB and one below AB.

Step 3: With the same radius and B as center draw two arcs to cut the arcs drawn in step 2. Mark the points of intersection of the arcs as C and D.

Step 4: Join C and D. CD will intersect AB. Mark the point of intersection as O CD is the required perpendicular bisector of AB.
Measure ∠AOC. Measure the length of AO and OB. What do you observe?

(i) Since the angles are linear pair, \(\angle\)ACD + \(\angle\)BCD = 180°
123° + \(\angle\)BCD = 180°
Subtracting 123° on both sides
123° + \(\angle\)BCD – 123° = 180° – 123°
\(\angle\)BCD = 57°
(ii) Since the angles are linear pair, \(\angle\)LNO + \(\angle\)MNO = 180°
46° + \(\angle\)MNO = 180°
Subtracting 46° on both sides
46° + \(\angle\)MNO – 46° = 180° – 46°
\(\angle\)MNO = 134°

We know that \(\angle\)POR = \(\angle\)POQ + \(\angle\)QOR
62° = 23° + \(\angle\)QOR
Subtracting 23° on both sides
62° – 23° = 23° + \(\angle\)QOR – 23°
\(\angle\)QOR = 39°

The two real - life situations where we use parallel lines are
(i) Railway tracks
(ii) Edges of a door frame

(i)  x = 125° (vertically opposite angles)
(ii)  y + 125 = 180⁰ (linear pair of angles)
y = 180° - 125°
y = 55⁰
∴ x^o = 125°; y^o= 55^o

From \(\angle \mathrm{PQR}=\angle \mathrm{SOP}\) (vertically opposite angels)
105 = x
(ie) x = 105^o

(i) any two pairs of adjacent angles are
(a) ∠PQT and ∠TOS 
(b) ∠PQU and ∠RQU 
(ii) two pairs of vertically opposite angles are
(a) ∠PQT and ∠RQU 
(b) ∠RQT and ∠PQU 

Let the three angles be x, 4x and 7x .
The sum of three angles at a points is supplementary
∴ x + 4x + 7x = 360°
12x = 360°
x^o = \(\frac{{360}^o}{12}\)
x = 30°
If x = 30^o then
4x = 4 x 30° = 120°
7x = 7 x 30° = 210°.
∴ The three angles are 30°, 120° and 210°.

From the figure
∴ ∠JlL = ∠JIK + ∠KIL
= 38° + 27°
= 65°
∴ ∠JlL = 65°

Step 1: Draw the given angle ∠ABC with the measure 80° using protractor.

Step 2: With B as center and convenient radius, draw an arc to cut BA and BC. Mark the points of intersection as E on BA and F on BC.

Step 3: With the same radius and E as center, draw an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc.

Step 4: Mark the point of intersection as G. Draw a ray BX through G. BG is the required bisector of the given angle ∠ABC

Construction:
Step 1:

Drawn a line. Marked two points A and B on it so that AB = 8 cm
Step2:

Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above AB and one below AB
Step3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2. Marked the points of intersection of the arcs as C and D.
Step4:

Joined C and D, CD intersect AB. Marked the point of intersection as 'O'. CD is the required perpendicular bisector of AB.
(b) 7 cm
Step 1:

Drawn a line and marked points A and B on it so that AB = 7cm.
Step 2:

Using compass with A as, centre and radius more than half of the length of AB drawn two arcs of same length one above AB and one below AB.
Step 3: With the same radius and B as centre drawn two arcs to cut the already drawn arcs in step 2. Marked the intersection of the arcs as C and D.
Step 4:

Joined C and D, CD is the required perpendicular bisector of AB.
(c) 5.6 cm.
Construction:

Step 1: Drawn a line and marked two points A and B on it so that AB = 5.6 cm
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length, one above AB and one below AB
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the arcs as C and D
Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as 'O' CD is the required perpendicular bisector of AB.
(d) 10.4 cm
Construction :

Step 1: Drawn a line and marked two points A and B on it so that AB = 10A cm.
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of same length one above AB and one below AB
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the arcs as C and D.
Step 4: Joined C and D. CD intersects AB. Marked the points of intersection as 0 I CD is the required perpendicular bisector
(e) 58 mm
Construction :

Step 1: Drawn a line. Marked two points A and B on it so that AB = 5.8 cm = 58 mm.
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above AB and one below AB.
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs of drawn in step 2. Marked the points of intersection of the arcs as C and D
Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as O. CD is the required perpendicular bisector.

(i) The angle that corresponds to \(\angle\)1 is \(\angle\)5
(ii) The angle that is alternate interior to \(\angle\)3 is \(\angle\)5
(iii) The angle that is alternate exterior to \(\angle\)8 is \(\angle\)2
(iv) The angle that corresponds to \(\angle\)8 is \(\angle\)4
(v) The angle that is alternate exterior to \(\angle\)7 is \(\angle\)1
(vi) The angle that is alternate interior to \(\angle\)6 is \(\angle\)4